Announce of Looksery Cup 2015-B. Looksery Party

B. Looksery Party

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The Looksery company, consisting of n staff members, is planning another big party. Every employee has his phone number and the phone numbers of his friends in the phone book. Everyone who comes to the party, sends messages to his contacts about how cool it is. At the same time everyone is trying to spend as much time on the fun as possible, so they send messages to everyone without special thinking, moreover, each person even sends a message to himself or herself.

Igor and Max, Looksery developers, started a dispute on how many messages each person gets. Igor indicates n numbers, the i-th of which indicates how many messages, in his view, the i-th employee is going to take. If Igor guesses correctly at least one of these numbers, he wins, otherwise Max wins.

You support Max in this debate, so you need, given the contact lists of the employees, to determine whether there is a situation where Igor loses. Specifically, you need to determine which employees should come to the party, and which should not, so after all the visitors send messages to their contacts, each employee received a number of messages that is different from what Igor stated.

Input

The first line contains a single integer n (1 ≤ n ≤ 100) — the number of employees of company Looksery.

Next n lines contain the description of the contact lists of the employees. The i-th of these lines contains a string of length n, consisting of digits zero and one, specifying the contact list of the i-th employee. If the j-th character of the i-th string equals 1, then the j-th employee is in the i-th employee's contact list, otherwise he isn't. It is guaranteed that the i-th character of the i-th line is always equal to 1.

The last line contains n space-separated integers: a1, a2, ..., an (0 ≤ ai ≤ n), where ai represents the number of messages that the i-th employee should get according to Igor.

Output

In the first line print a single integer m — the number of employees who should come to the party so that Igor loses the dispute.

In the second line print m space-separated integers — the numbers of these employees in an arbitrary order.

If Igor wins the dispute in any case, print -1.

If there are multiple possible solutions, print any of them.

Sample test(s)

input

31010100010 1 2

output

11 

input

111

output

0

input

411110101111000011 0 1 0

output

41 2 3 4 

Note

In the first sample Igor supposes that the first employee will receive 0 messages. Since he isn't contained in any other contact list he must come to the party in order to receive one message from himself. If he is the only who come to the party then he will receive 1 message, the second employee will receive 0 messages and the third will also receive 1 message. Thereby Igor won't guess any number.

In the second sample if the single employee comes to the party he receives 1 message and Igor wins, so he shouldn't do it.

In the third sample the first employee will receive 2 messages, the second — 3, the third — 2, the fourth — 3.

B. Looksery Party

Author: Igor_Kudryashov

In any cases there is such set of people that if they come on party and send messages to their contacts then each employee receives the number of messages that is different from what Igor pointed. Let's show how to build such set. There are 2 cases.

There are no zeros among Igor's numbers. So if nobody comes on party then each employee receives 0 messages and, therefore, the desired set is empty.

There is at least one zero. Suppose Igor thinks that i-th employee will receive 0 messages. Then we should add i-th employee in the desired set. He will send messages to his contacts and will receive 1 message from himself. If we add other employees in the desired set then the number of messages that i-th employee will receive will not decrease so we can remove him from considering. Igor pointed some numbers for people from contact list of i-th employee and because they have already received one message we need to decrease these numbers by one. After that we can consider the same problem but with number of employees equals to n - 1. If the remaining number of employees is equal to 0 then the desired set is built.

 1.如果猜测的数字没有0，那么一个人也不选；

2.如果猜测的数字有至少一个0，那么选他，他将收到自己的一封短信，因此我们以后可以不再考虑他，同时将他发送给的人的预测数减1。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<vector>using namespace std;const int INF=500;int n;int g[103][103];string ch[103];struct Node{    int x, cnt;    bool operator < (const Node & x) const {        return cnt<x.cnt;    }}num[103];//int num[103];vector<int> zero;vector<int> ans;bool f(int n){    sort(num, num+n+1);    if(num[0].cnt<0) {        num[0].cnt=INF;        return true;    }    if(num[0].cnt!=0) return false;    for(int i=1;i<n;i++){        if(g[num[0].x][num[i].x]) num[i].cnt--;    }    ans.push_back(num[0].x);    num[0].cnt=INF;    return true;}void solve(){    for(int i=n;i>=0;i--){        if(!f(i)) break;    }    if(ans.size()==0) cout<<"0"<<endl;    else {        cout<<ans.size()<<endl;        for(int i=0;i<ans.size();i++){            if(i) cout<<' ';            cout<<ans[i]+1;        }        cout<<endl;    }}int main(){    cin>>n;    for(int i=0;i<n;i++){        cin>>ch[i];    }    for(int i=0;i<n;i++)        for(int j=0;j<n;j++)            g[i][j]=ch[i][j]-'0';    for(int i=0;i<n;i++) {        cin>>num[i].cnt;        num[i].x=i;    }    num[n].cnt=INF;    solve();}