The Way to Home (模拟)

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本文介绍了一个经典的算法问题——青蛙如何在限定条件下用最少的跳跃次数从起点1抵达终点n。通过分析给定的01字符串，确定每个位置是否有可用的落脚点，并实现了一种有效的算法来解决此问题。

The Way to Home

A frog lives on the axis Ox and needs to reach home which is in the point n. She starts from the point 1. The frog can jump to the right at a distance not more than d. So, after she jumped from the point x she can reach the point x + a, where a is an integer from 1 to d.

For each point from 1 to n is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and n.

Determine the minimal number of jumps that the frog needs to reach home which is in the point n from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1.

Input

The first line contains two integers n and d (2 ≤ n ≤ 100, 1 ≤ d ≤ n - 1) — the point, which the frog wants to reach, and the maximal length of the frog jump.

The second line contains a string s of length n, consisting of zeros and ones. If a character of the string s equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string s equal to one.

Output

If the frog can not reach the home, print -1.

In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point n from the point 1.

Examples

Input

8 4
10010101

Output

Input

4 2
1001

Output

-1

Input

8 4
11100101

Output

Input

12 3
101111100101

Output

Note

In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four).

In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two.

给定n，d，两个整数，然后一个01字符串，n是这个字符串的长度，青蛙只能向右跳，d是每次它向右跳的最大步数，要求每次只能跳到1上，问青蛙从1跳到n最少跳几次

我的思路是从1起点开始，先加上最大跳跃步数，然后从这个下标往回找找到1，更新起点，答案加一，否则不能到达输出-1

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main(){
    char s[109];
    int n,d;
    scanf("%d%d",&n,&d);
    getchar();
    for(int i = 1; i <= n; i++){
        scanf("%c",&s[i]);
    }
    int i = 1;
    int ans = 0;
    while(i <= n){//起点
        int j = i + d;//每次从起点跳最远到达的下标
        if(j > n){//如果从起点跳的最远下标超过了n肯定可达并结束
            ans++;
            break;
        }
        int flag = 0;
        while(j > i){
            if(s[j] == '1'){//找到第一个1停止
                ans++;
                flag = 1;
                i = j;
                break;
            }
            j--;
        }
        if(i == n) break;//如果恰好到了n退出循环
        if(!flag){//如果没有1说明不能到达
            printf("-1\n");
            return 0;
        }
    }
    printf("%d\n",ans);
    return 0;
}