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Look-and-say sequence is a sequence of integers as the following:

D, D1, D111, D113, D11231, D112213111, ...

where D is in [0, 9] except 1. The (n+1)st number is a kind of description of the nth number. For example, the 2nd number means that there is one D in the 1st number, and hence it is D1; the 2nd number consists of one D (corresponding to D1) and one 1 (corresponding to 11), therefore the 3rd number is D111; or since the 4th number is D113, it consists of one D, two 1's, and one 3, so the next number must be D11231. This definition works for D = 1 as well. Now you are supposed to calculate the Nth number in a look-and-say sequence of a given digit D.

Input Specification:

Each input file contains one test case, which gives D (in [0, 9]) and a positive integer N (≤ 40), separated by a space.

Output Specification:

Print in a line the Nth number in a look-and-say sequence of D.

Sample Input:

1 8

Sample Output:

1123123111

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <ctime>
#include <cctype>
#include <bitset>
#include <utility>
#include <sstream>
#include <complex>
#include <iomanip>
#include<climits>//INT_MAX

//#include<bits/stdc++.h>
#define PP pair<ll,int>
#define inf 0x3f3f3f3f
#define llinf 0x3f3f3f3f3f3f3f3fll
#define dinf 1000000000000.0
#define PI 3.1415926
typedef long long ll;
using namespace std;
int const mod=1e9+7;
const int maxn=1e5+10;
int d, n;
int main() {
    scanf("%d %d", &d, &n);
    vector<int> V;
    V.push_back(d);
    for (int i=1; i<n; i++) {
        vector<int> ls;
        int v=V[0];
        int ct=0;
        for (int j=0; j<V.size(); j++) {
            if (V[j]==v)
                    ct++;
            else{
                ls.push_back(v);
                ls.push_back(ct);
                ct=1;
                v=V[j];
            }
            if (j==V.size()-1) {
                ls.push_back(v);
                ls.push_back(ct);
            }
        }
        V=ls;
    }
    for (int i=0; i<V.size(); i++)
        printf("%d", V[i]);
    return 0;
}