Problem:
﻿Score: 300 points

Problem Statement

Mr. Infinity has a string S consisting of digits from 1 to 9. Each time the date changes, this string changes as follows:

Each occurrence of 2 in S is replaced with 22. Similarly, each 3 becomes 333, 4 becomes 4444, 5 becomes 55555, 6 becomes 666666, 7 becomes 7777777, 8 becomes 88888888 and 9 becomes 999999999. 1 remains as 1.

For example, if S is 1324, it becomes 1333224444 the next day, and it becomes 133333333322224444444444444444 the day after next.
You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days?

Constraints

S is a string of length between 1 and 100 (inclusive).

K is an integer between 1 and 10^{18} (inclusive).

The length of the string after 5 \times 10^{15} days is at least K.

Input

Input is given from Standard Input in the following format:

S
K

Output

Print the K-th character from the left in Mr. Infinity's string after 5 \times 10^{15} days.

Sample Input 1

1214
4

Sample Output 1

2

The string S changes as follows:

Now: 1214

After one day: 12214444

After two days: 1222214444444444444444

After three days: 12222222214444444444444444444444444444444444444444444444444444444444444444

The first five characters in the string after 5 \times 10^{15} days is 12222. As K=4, we should print the fourth character, 2.

Sample Input 2

3
157

Sample Output 2

3

The initial string is 3. The string after 5 \times 10^{15} days consists only of 3.

Sample Input 3

299792458
9460730472580800

Sample Output 3

2

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;


//repetition
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define euc_dis(x, y) sqrt(x*x + y*y)

#define pb push_back
#define mp make_pair
#define INF 999999999
#define MOD 1000000007
#define sp ' '
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()

//typedef
typedef long long ll;
typedef pair<int, int> pint;
typedef map<int, int> mint;
typedef vector<int> vint;
typedef vector<char> vch;
typedef vector<long long> vll;
typedef vector<string> vstr;


int mod(int a, int b){return (a%b+b)%b;}
int gcd(int a,int b){return b?gcd(b,a%b):a;}
void Yes(){cout << "Yes" << endl;}
void No(){cout << "No" << endl;}
void Judge(bool b){b?Yes():No();}
void YES(){cout << "YES" << endl;}
void NO(){cout << "NO" << endl;}
void JUDGE(bool b){b?YES():NO();}
ll powMod(ll a, ll b){ll ans=1; rep(i, b){ans=ans*a%MOD;} return ans;}

template<typename T>
void ppp(T n){cout << n << endl;}
template<typename T>
void dpp(T n){cerr << n << endl;}

const int dx[4]={1,0,-1,0};
const int dy[4]={0,1,0,-1};
const int ddx[8] = {1,0,-1,0,1,-1,-1,1};
const int ddy[8] = {0,1,0,-1,1,1,-1,-1};




template < typename T > std::string to_string( const T& n )
{
    std::ostringstream stm ;
    stm << n ;
    return stm.str() ;
}


vector<string> split(const string &str, char sep)
{
    vector<string> v;
    stringstream ss(str);
    string buffer;
    while( getline(ss, buffer, sep) ) {
        v.pb(buffer);
    }
    return v;
}




int main(){

    cin.tie(0);
    ios::sync_with_stdio(false);

    string s;
    ll k;
    cin >> s >> k;
    ll five = 5000000000000000;
    vll length(s.size());

    ll one_cnt=0;
    rep(i, s.size()){
      if(s[i]=='1')
        one_cnt++;
    }

    if(s[0] == '1' and k <= one_cnt){
      ppp(1);
    }
    else if(s[0] == '1' and k > one_cnt){
      ppp(s[1]);
    }
    else{
      ppp(s[0]);
    }

}