CodeForces 729D Sea Battle

题目衔接：http://codeforces.com/problemset/problem/729/D

D. Sea Battle

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Galya is playing one-dimensional Sea Battle on a 1 × n grid. In this game a ships are placed on the grid. Each of the ships consists of bconsecutive cells. No cell can be part of two ships, however, the ships can touch each other.

Galya doesn't know the ships location. She can shoot to some cells and after each shot she is told if that cell was a part of some ship (this case is called "hit") or not (this case is called "miss").

Galya has already made k shots, all of them were misses.

Your task is to calculate the minimum number of cells such that if Galya shoot at all of them, she would hit at least one ship.

It is guaranteed that there is at least one valid ships placement.

Input

The first line contains four positive integers n, a, b, k (1 ≤ n ≤ 2·105, 1 ≤ a, b ≤ n, 0 ≤ k ≤ n - 1) — the length of the grid, the number of ships on the grid, the length of each ship and the number of shots Galya has already made.

The second line contains a string of length n, consisting of zeros and ones. If the i-th character is one, Galya has already made a shot to this cell. Otherwise, she hasn't. It is guaranteed that there are exactly k ones in this string.

Output

In the first line print the minimum number of cells such that if Galya shoot at all of them, she would hit at least one ship.

In the second line print the cells Galya should shoot at.

Each cell should be printed exactly once. You can print the cells in arbitrary order. The cells are numbered from 1 to n, starting from the left.

If there are multiple answers, you can print any of them.

Examples

input

Copy

5 1 2 1
00100

output

Copy

2
4 2

input

Copy

13 3 2 3
1000000010001

output

Copy

2
7 11

Note

There is one ship in the first sample. It can be either to the left or to the right from the shot Galya has already made (the "1" character). So, it is necessary to make two shots: one at the left part, and one at the right part.

题目大意：现在某人要找到一些船，给你一个已被某人射击的串，1代表已经射击0代表未射击，输入nabk，分别代表串长，船的数量，船的宽度，和射击的次数，现在问你有多少艘船，并输出船的位置

思路：直接遍历查找，遇到1或者0的个数为b即为一个船，存进数组最后输出即可

代码：

/*

*/
#include<map>
#include<set>
#include <vector>
#include<stack>
#include<queue>
#include<cmath>
#include<string>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll unsigned long long
#define inf 0x3f3f3f
#define esp 1e-8
#define bug {printf("mmp\n");}
#define mm(a,b) memset(a,b,sizeof(a))
#define T() int test,q=1;scanf("%d",&test); while(test--)
const int maxn=2e5+10;
const double pi=acos(-1.0);
const int N=201;
const int mod=1e9+7;

char s[maxn];
int aa[maxn];
int main()
{
    int n,a,b,k;
    cin>>n>>a>>b>>k;
    scanf("%s",s);
    int l=0,cnt=0;
    for(int i=0;i<n;i++)
    {
        if(s[i]=='0')
        {
            cnt++;
            if(cnt==b)
            {
                aa[l++]=i+1;
                cnt=0;
            }
        }
        else if(s[i]=='1')
        {
            cnt=0;
        }
    }
    printf("%d\n",l-a+1);
    printf("%d",aa[0]);
    for(int i=1;i<=l-a;i++)
    {
        printf(" %d",aa[i]);
    }
    printf("\n");
    return 0;
}