Problem:
Score : 200 points

Problem Statement

You are given a string S of length N consisting of lowercase English letters, and an integer K.
Print the string obtained by replacing every character in S that differs from the K-th character of S, with *.

Constraints

1 \leq K \leq N\leq 10

S is a string of length N consisting of lowercase English letters.

N and K are integers.

Input

Input is given from Standard Input in the following format:

N
S
K

Output

Print the string obtained by replacing every character in S that differs from the K-th character of S, with *.

Sample Input 1

5
error
2

Sample Output 1

*rr*r

The second character of S is r. When we replace every character in error that differs from r with *, we get the string *rr*r.

Sample Input 2

6
eleven
5

Sample Output 2

e*e*e*

Sample Input 3

9
education
7

Sample Output 3

******i**

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

#include<iostream>
#include<map>
#include<set>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
#include<limits>
#include<sstream> 
#include<unordered_map>      

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef int itn;
typedef vector<int> vi;
typedef vector<ll> vl;

const int inf = numeric_limits<int>::max() >> 2;
const ll linf = numeric_limits<ll>::max() >> 2;
const ull ulinf = numeric_limits<ull>::max() >> 2;
const double pi = acos(-1);
const int dx[4]={1,0,-1,0};
const int dy[4]={0,1,0,-1};
const int dx8[8]={-1,0,1,-1,1,-1,0,1};
const int dy8[8]={-1,-1,-1,0,0,1,1,1};

#define p_queue(i) priority_queue<i> 
#define rp_queue(i) priority_queue<i, vector<i>, greater<i>> 
#define umap(i,j) unordered_map<i,j>
#define P(p) cout<<(p)<<endl
#define PS(p) cout<<(p)<< " "
#define IN cin >> 
#define rep(i,m,n) for(int i = (m); i < (int)(n); i++)
#define rrep(i,m,n) for(int i = (m-1); i >= (int)(n); i--)
#define inrep(n,a) for(int i = 0; i < (int)(n); i++)cin >> a[i];
#define mod(i) ((i)%(ll)(1e9 + 7))
#define divm(a,b) (mod(a * modpow((ll)b,(ll)(1e9+5))))
#define rsort(a,b,c) sort(a, b, greater<c>())
#define vsort(v) sort(v.begin(), v.end())
#define rvsort(v,c) sort(v.begin(), v.end(), greater<c>()) 
#define ft first
#define sd second
#define pb push_back
#define it insert
#define sz(x) ((int)(x).size())
#define lb(a,n,k) (lower_bound(a,a+n,k) - a) 
#define vlb(a,k)  (lower_bound(a.begin(),a.end(),k) - a.begin())
#define ub(a,n,k) (upper_bound(a,a+n,k) - a) 
#define vub(a,k)  (upper_bound(a.begin(),a.end(),k) - a.begin())
#define YES cout<<"YES"<< endl
#define NO cout<<"NO"<<endl
#define Yes cout<<"Yes"<<endl
#define No cout<<"No"<<endl  
#define yes cout<<"yes"<<endl
#define no cout<<"no"<<endl
#define ret return

ll modpow(ll i,ll j){ ll tmp=1; while(j){ if(j%2)tmp=mod(tmp*i);i=mod(i*i);j/=2;}return tmp;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }

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

//ll ncr[100][100];
//ll nCr(int n, int r){if(n==r) ret ncr[n][r] = 1; if(r==0) ret ncr[n][r] = 1; if(r==1) ret ncr[n][r] = n;if(ncr[n][r]) ret ncr[n][r]; ret ncr[n][r] = nCr(n-1,r) + nCr(n-1,r-1);}

//ll npr[100][100]={};
//ll nPr(int n,int r){if(npr[n][r])ret npr[n][r];if(r==0)ret npr[n][r] = 1;if(r==1)ret npr[n][r] = n;ret npr[n][r] = n * nPr(n-1,r-1);}

//ll nHr(int n,int r){ret nCr(n+r-1,r);}

///////////////////////////////////////////////////////////////////////////


int main(){

    int n,k;
    string s;

    cin >> n >> s >> k;

    rep(i,0,n){
	if(s[k-1] != s[i])cout << "*";
	else cout << s[i];
    }
    cout << endl;
    return  0;
}