A string s of length n can be encrypted by the following algorithm:

  * iterate over all divisors of n in decreasing order (i.e. from n to 1), 
  * for each divisor d, reverse the substring s[1 ... d] (i.e. the substring which starts at position 1 and ends at position d). 



For example, the above algorithm applied to the string s="codeforces" leads to the following changes: "codeforces" → "secrofedoc" → "orcesfedoc" → "rocesfedoc" → "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1).

You are given the encrypted string t. Your task is to decrypt this string, i.e., to find a string s such that the above algorithm results in string t. It can be proven that this string s always exists and is unique.

Input

The first line of input consists of a single integer n (1 ≤ n ≤ 100) — the length of the string t. The second line of input consists of the string t. The length of t is n, and it consists only of lowercase Latin letters.

Output

Print a string s such that the above algorithm results in t.

Examples

Input

10
rocesfedoc


Output

codeforces


Input

16
plmaetwoxesisiht


Output

thisisexampletwo


Input

1
z


Output

z

Note

The first example is described in the problem statement.
var n = parseInt(readline(), 10);
var s = readline();
var divisor = [];

for(var i=1; i*i<=n; i++) {
    if(n%i === 0) {
        if(i === n/i) {
            divisor.push(i);
            
        } else {
            divisor.push(i);
            divisor.push(n/i);
            
        }
    }
}

divisor.sort((a,b) => a-b);

function reverseIt(str) {
    arr = str.split('');
    
    for(var i=arr.length-1; i>=(arr.length/2); i--) {
        var temp = arr[i];
        arr[i] = arr[arr.length-i-1];
        arr[arr.length-i-1] = temp;
    }
    
    return arr.join('');
}

for(var i=0; i<divisor.length; i++) {
    var temp = s.substring(0, divisor[i]);
    s = s.replace(temp, reverseIt(temp));
        
}

print(s);
    						    				  		 							