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.
import java.util.*;

public class Check2 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        String s = sc.next();
        for(int i=1;i<=n;i++){
            if(n%i==0){
                String s1=s.substring(0,i);
                String s2=s.substring(i);
                StringBuilder stringBuilder=new StringBuilder(s1);
                s=stringBuilder.reverse()+s2;
                //System.out.println(s+" "+i);
            }
        }
        System.out.println(s);





    }


    public static long power(long a, long b, long c) {
        long ans = 1;
        while (b != 0) {
            if (b % 2 == 1) {
                ans = ans * a;
                ans %= c;
            }
            a = a * a;
            a %= c;
            b /= 2;
        }
        return ans;
    }

    public static long power1(long a, long b, long c) {
        long ans = 1;
        while (b != 0) {
            if (b % 2 == 1) {
                ans = multiply(ans, a, c);
            }
            a = multiply(a, a, c);
            b /= 2;

        }
        return ans;

    }

    public static long multiply(long a, long b, long c) {
        long res = 0;
        a %= c;
        while (b > 0) {
            if (b % 2 == 1) {
                res = (res + a) % c;
            }
            a = (a + a) % c;
            b /= 2;
        }

        return res % c;

    }

    public static long totient(long n) {
        long result = n;
        for (long i = 2; i * i <= n; i++) {
            if (n % i == 0) {
                //sum=sum+2*i;
                while (n % i == 0) {
                    n /= i;
                    // sum=sum+n;
                }
                result -= result / i;

            }
        }
        if (n > 1) {
            result -= result / n;

        }
        return result;
    }


    public static long gcd(long a, long b) {
        if (b == 0) {
            return a;
        } else {
            return gcd(b, a % b);
        }
    }

    public static boolean[] primes(int n) {
        boolean[] p = new boolean[n + 1];
        p[0] = false;
        p[1] = false;
        for (int i = 2; i <= n; i++) {
            p[i] = true;
        }

        for (int i = 2; i * i <= n; i++) {
            if (p[i]) {
                for (int j = i * i; j <= n; j += i) {
                    p[j] = false;
                }
            }

        }
        return p;
    }

    static int val(char c) {
        if (c >= '0' && c <= '9')
            return (int) c - '0';
        else
            return (int) c - 'A' + 10;
    }

    // Function to convert a
// number from given base
// 'b' to decimal
    static int toDeci(String str,
                      int base) {
        int len = str.length();
        int power = 1; // Initialize
        // power of base
        int num = 0; // Initialize result
        int i;

        // Decimal equivalent is
        // str[len-1]*1 + str[len-2] *
        // base + str[len-3]*(base^2) + ...
        for (i = len - 1; i >= 0; i--) {
            // A digit in input number
            // must be less than
            // number's base
            if (val(str.charAt(i)) >= base) {
                System.out.println("Invalid Number");
                return -1;
            }

            num += val(str.charAt(i)) * power;
            power = power * base;
        }

        return num;
    }
}