A reversible prime in any number system is a prime whose "reverse" in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime.

Now given any two positive integers N (<10​5​​) and D (1<D≤10), you are supposed to tell if N is a reversible prime with radix D.

Input Specification:

The input file consists of several test cases. Each case occupies a line which contains two integers N and D. The input is finished by a negative N.

Output Specification:

For each test case, print in one line Yes if N is a reversible prime with radix D, or No if not.

Sample Input:

73 10

23 2

23 10

-2

Sample Output:

Yes

Yes

No

#include <stdio.h>

#include <stdlib.h>

#include <math.h>

#include <algorithm>

#include <iostream>

#include <string.h>

#include <queue>

#include <string>

using namespace std;

const int maxn = ;

int n, d;

int to_10(int i,int radix) {

    int res = ,exp=;

    while (i != ) {

        res += i%*pow(radix, exp);

        exp++;

        i /= ;

    }

    return res;

}

bool is_prime(int i) {

    if (i ==  || i == )return false;

    if (i ==  || i == )return true;

    for (int j = ; j*j <= i; j++) {

        if (i%j == )return false;

    }

    return true;

}

string to_s(int i,int radix) {

    string s = "";

    do {

        s += '' + i % radix;

        i /= radix;

    } while (i != );

    reverse(s.begin(), s.end());

    return s;

}

int to_num(string s) {

    int res = ;

    for (int i = s.length()-; i >=; i--) {

        res = res *  + s[i] - '';

    }

    return res;

}

int main() {

    while (true) {

        scanf("%d", &n);

        if (n < )break;

        else {

            scanf("%d", &d);

            if (is_prime(n) && is_prime(to_10(to_num(to_s(n,d)), d))) printf("Yes\n");

            else printf("No\n");

        }

    }

    system("pause");

    return ;

}