Consider an ordered set S of strings of N (1 <= N <= 31) bits. Bits, of course, are either 0 or 1.

This set of strings is interesting because it is ordered and contains all possible strings of length N that have L (1 <= L <= N) or fewer bits that are `1'.

Your task is to read a number I (1 <= I <= sizeof(S)) from the input and print the Ith element of the ordered set for N bits with no more than L bits that are `1'.

A single line with three space separated integers: N, L, and I.

A single line containing the integer that represents the Ith element from the order set, as described.

 #include<bits/stdc++.h>

 using namespace std;

 #define ll long long

 ll dp[][],I;

 int main()

 {

     int N,L;

     cin>>N>>L>>I;

     dp[][]=;

     for(int i=;i<=N;i++)

         for(int j=;j<=L;j++)

             if(j==)dp[i][j]=dp[i-][j];

             else if(j>i)dp[i][j]=dp[i][i];

             else if(j==i)dp[i][j]=dp[i][j-]+;

             else dp[i][j]=dp[i-][j]+dp[i-][j-];

     for(int i=N-;i>=;i--)

         if(dp[i][L]<I){cout<<"";I-=dp[i][L--];}

         else cout<<"";

     if(I!=)cout<<"";

     else cout<<"";

     return ;

 }