Codeforces650B Image Preview

B. Image Preview

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya's telephone contains n photos. Photo number 1 is currently opened on the phone. It is allowed to move left and right to the adjacent photo by swiping finger over the screen. If you swipe left from the first photo, you reach photo n. Similarly, by swiping right from the last photo you reach photo1. It takes a seconds to swipe from photo to adjacent.

For each photo it is known which orientation is intended for it — horizontal or vertical. Phone is in the vertical orientation andcan't be rotated. It takes b second to change orientation of the photo.

Vasya has T seconds to watch photos. He want to watch as many photos as possible. If Vasya opens the photo for the first time, he spends1 second to notice all details in it. If photo is in the wrong orientation, he spendsb seconds on rotating it before watching it. If Vasya has already opened the photo, he just skips it (so he doesn't spend any time for watching it or for changing its orientation). It is not allowed to skip unseen photos.

Help Vasya find the maximum number of photos he is able to watch during T seconds.

Input

The first line of the input contains 4 integers n, a, b, T (1 ≤ n ≤ 5·10⁵,1 ≤ a, b ≤ 1000, 1 ≤ T ≤ 10⁹) — the number of photos, time to move from a photo to adjacent, time to change orientation of a photo and time Vasya can spend for watching photo.

Second line of the input contains a string of length n containing symbols 'w' and 'h'.

If the i-th position of a string contains 'w', then the photoi should be seen in the horizontal orientation.

If the i-th position of a string contains 'h', then the photoi should be seen in vertical orientation.

Output

Output the only integer, the maximum number of photos Vasya is able to watch during thoseT seconds.

Examples

Input

4 2 3 10wwhw

Output

2

Input

5 2 4 13hhwhh

Output

4

Input

5 2 4 1000hhwhh

Output

5

Input

3 1 100 10whw

Output

0

Note

In the first sample test you can rotate the first photo (3 seconds), watch the first photo (1 seconds), move left (2 second), rotate fourth photo (3 seconds), watch fourth photo (1 second). The whole process takes exactly 10 seconds.

Note that in the last sample test the time is not enough even to watch the first photo, also you can't skip it.

题意：Vasya从第一张照片开始看，可以向左翻或者向右，第1张左边是第n张，第n张右边是第一张，翻一次要a时间，照片有'w'和'h'两种摆放，‘w’的照片要耗费b时间翻转，看一张照片要1时间，一共有T时间，问最多可以看多少张照片？

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define ll long long int#define maxn 500005int sum[2*maxn];char s[maxn];int main(){    int n, a, b, T, i, h, r, ans, ss;    scanf("%d %d %d %d", &n, &a, &b, &T);    scanf("%s", s);    ss = 0;    for(i = 0;i < n;i++){        if(s[i] == 'w') sum[i] = sum[i+n] = b + 1;        else sum[i] = sum[i+n] = 1;        ss += sum[i];    }    ss -= sum[0];    h = 1, r = n, ans = 0;    while(h<=n&&r<2*n){        ss += sum[r++];        //    维持n以内||先右后左    先左后右        while(r-h>n||ss+(r-h-1+min(r-n-1, n-h))*a>T){            ss -= sum[h++];        }        ans = max(ans, r-h);    }    printf("%d\n", ans);}