Codeforces Round #345 (Div. 2) D. Image Preview【二分】

D. 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 photo 1. 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 and can'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 spends 1 second to notice all details in it. If photo is in the wrong orientation, he spends b 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 photo i should be seen in the horizontal orientation.

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

Output

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

Examples

input

4 2 3 10
wwhw

output

2

input

5 2 4 13
hhwhh

output

4

input

5 2 4 1000
hhwhh

output

5

input

3 1 100 10
whw

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：你的手机上有n张照片，你可以向左浏览，也可以向右浏览，观看每张照片需要时间1秒。每张照片都有两个方向，水平或者竖直方向，手机只能观看竖直方向的图片，给你切换照片需要的时间a秒，改变照片方向的时间b秒，以及你拥有的时间T秒。输入n张照片的方向，输出在T秒内你能观看最多的照片数目。

求出v[i]=v[i-1]+1+(s[i]=='w'?b:0);再进行二分

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
const int maxn=5e5+5;
int n,a,b,T,v[maxn];
char s[maxn];
bool C(int mid)
{
    int sum1,sum2;
    for(int i=1;i<=mid;i++){
        sum1=v[i]+(v[n]-v[n-mid+i]);  //左移mid-i位总和
        sum2=min((i-1)*a*2+(mid-i)*a,(i-1)*a+(mid-i)*a*2);//求出(向右i-1折返+向左mid-i)与(向左mid-i折返+向右i-1)的最小值
        if(sum1+sum2<=T){
            return true;
        }
    }
    return false;
}
int main()
{
    cin>>n>>a>>b>>T;
    cin>>s+1;
    for(int i=1;i<=n;i++){
        v[i]=v[i-1]+1+(s[i]=='w'?b:0);
    }
    int l=0,r=n,mid;
    while(l<r){
        mid=(l+r+1)/2;
        if(C(mid)){
            l=mid;
        }
        else{
            r=mid-1;
        }
    }
    cout<<r<<endl;

    return 0;
}