Yltx Prime Pair Check

A yltx prime pair is defined as follows: Given two integers \(x\) and \(y\), if the expression \[ x \times y - 3\times (x-y) \] results in a prime number, then the pair \((x,y)\) is called a yltx prime pair.

You are given \(T\) pairs generated using a seed \((x_0,y_0)\). The data are generated by the following procedure:

// C++ pseudocode for data generation
#include
using namespace std;
int T, x_0, y_0;
int main() {
    scanf("%d%d%d", &T, &x_0, &y_0);
    while(T--) {
        x_0 = ((7*x_0+13) ^ (x_0/13-7));
        y_0 = ((7*y_0+13) ^ (y_0/13-7));
        int x = ((x_0 % 10000 + 10000) % 10000) + 1;
        int y = ((y_0 % 10000 + 10000) % 10000) + 1;
        // (x, y) is one test case
    }
    return 0;
}

Your task is to simulate the above data generation process and for each generated pair \((x,y)\), check whether it is a yltx prime pair. In other words, for each pair, compute \[ E = x \times y - 3\times (x-y), \] and if \(E\) is a prime number (note that a prime is an integer greater than 1 that has no positive divisors other than 1 and itself), then the answer for that test case is "Yes"; otherwise, it is "No".

inputFormat

The first line of input contains three integers: \(T\), \(x_0\), and \(y_0\). These represent the number of test cases and the seed values, respectively.

Data for each test case are generated using the following procedure:

while(T--) {
    x_0 = ((7*x_0+13) ^ (x_0/13-7));
    y_0 = ((7*y_0+13) ^ (y_0/13-7));
    int x = ((x_0 % 10000 + 10000) % 10000) + 1;
    int y = ((y_0 % 10000 + 10000) % 10000) + 1;
    // use x and y
}

outputFormat

For each test case, output a single line containing either Yes if the corresponding pair \((x,y)\) is a yltx prime pair, or No otherwise.

sample

3 1 1

No
No
No