C. Xenia and Weights

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play with scales and weights a little. For this, she puts weights on the scalepans, one by one. The first weight goes on the left scalepan, the second weight goes on the right scalepan, the third one goes on the left scalepan, the fourth one goes on the right scalepan and so on. Xenia wants to put the total of m weights on the scalepans.
Simply putting weights on the scales is not interesting, so Xenia has set some rules. First, she does not put on the scales two consecutive weights of the same weight. That is, the weight that goes i-th should be different from the (i + 1)-th weight for any i (1 ≤ i < m). Second, every time Xenia puts a weight on some scalepan, she wants this scalepan to outweigh the other one. That is, the sum of the weights on the corresponding scalepan must be strictly greater than the sum on the other pan.
You are given all types of weights available for Xenia. You can assume that the girl has an infinite number of weights of each specified type. Your task is to help Xenia lay m weights on ​​the scales or to say that it can't be done.

Input

The first line contains a string consisting of exactly ten zeroes and ones: the i-th (i ≥ 1) character in the line equals "1" if Xenia has i kilo weights, otherwise the character equals "0". The second line contains integer m (1 ≤ m ≤ 1000).

Output

In the first line print "YES", if there is a way to put m weights on the scales by all rules. Otherwise, print in the first line "NO". If you can put m weights on the scales, then print in the next line m integers — the weights' weights in the order you put them on the scales.
If there are multiple solutions, you can print any of them.

Sample test(s)

Input

0000000101
3

Output

YES
8 10 8

Input

1000000000
2

Output

NO

This was a very nice dfs type dp question accompanied by few observations.
First of all we notice that if the difference at any point of time is greater than 10 then the state cannot be
continued. 
Suppose the weight configuration in the pan is (16,20,10) where 16 is the sum of weight in left pan, 20 is the sum 
of weight in right pan and last used weight is 10 this can be simplified to the state as (4,10,number of steps) 
where 4 is the difference and 10 is the last used weight in right pan. So we keep a state 
dp[difference][last used weight][number of steps]. And whenever we assign a new weight we want the present weight
to be greater than the difference. 

The code for clarity

#include<bits/stdc++.h>
using namespace std;
int dp[30][12][1010]  ;// dp[diiference][last][steps]
char h[12];
int arr[1010];
bool f=0;
int m;
bool solve(int diff, int last, int steps,int fill[],int pos)
{
  // cout<<diff<<" "<<last<<" "<<steps<<endl;
   if(steps==0)
   {
   // cout<<"return "<<endl;
      if(f==0)
      {
       cout<<"YES"<<endl;
          for(int i=0;i<m;i++)
    cout<<fill[i]<<" ";
    f=1; 
    }
   return 1;
      
   }
   if(dp[diff][last][steps]!=-1)
   {
    //cout<<"ret "<<endl;
   return dp[diff][last][steps];
      }
   bool ret=0;
   for(int i=0;i<10;i++)
   {
   // cout<<"enter "<<endl;
            if(h[i]=='1' && i+1!=last && i+1>diff)
            {
    //         cout<<"call solve "<<endl;
             fill[pos]=i+1;
                      ret=solve(i+1-diff,i+1,steps-1,fill,pos+1); 
    }
    if(ret==1)
    break;
   }
   dp[diff][last][steps]=ret;
   return ret;
}
int main()
{
 //int m;
 string s;
 cin>>s;
 for(int i=0;i<10;i++)
 {
    h[i]=s[i];
 }
 memset(dp,-1,sizeof(dp));
 cin>>m;
 bool ans=0;
 ans=solve(0,0,m,arr,0);
 if(!ans)
 {
   cout<<"NO"<<endl;
   return 0;
 }
 
 return 0;
}