Takahashi is doing a research on sets of points in a plane. Takahashi thinks a set S of points in a coordinate plane is a good set when S satisfies both of the following conditions:

* The distance between any two points in S is not \sqrt{D_1}.
* The distance between any two points in S is not \sqrt{D_2}.



Here, D_1 and D_2 are positive integer constants that Takahashi specified.

Let X be a set of points (i,j) on a coordinate plane where i and j are integers and satisfy 0 ≤ i,j < 2N.

Takahashi has proved that, for any choice of D_1 and D_2, there exists a way to choose N^2 points from X so that the chosen points form a good set. However, he does not know the specific way to choose such points to form a good set. Find a subset of X whose size is N^2 that forms a good set.

Constraints

* 1 ≤ N ≤ 300
* 1 ≤ D_1 ≤ 2×10^5
* 1 ≤ D_2 ≤ 2×10^5
* All values in the input are integers.

Input

Input is given from Standard Input in the following format:


N D_1 D_2


Output

Print N^2 distinct points that satisfy the condition in the following format:


x_1 y_1
x_2 y_2
:
x_{N^2} y_{N^2}


Here, (x_i,y_i) represents the i-th chosen point. 0 ≤ x_i,y_i < 2N must hold, and they must be integers. The chosen points may be printed in any order. In case there are multiple possible solutions, you can output any.

Examples

Input

2 1 2


Output

0 0
0 2
2 0
2 2


Input

3 1 5


Output

0 0
0 2
0 4
1 1
1 3
1 5
2 0
2 2
2 4
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

bool A[777][777];

int main() {
    int n, a, b;
    scanf("%d%d%d", &n, &a, &b);
    n *= 2;
    memset(A, 1, sizeof A);
    int tmpa = 0;
    while(a % (1 << (tmpa + 1)) == 0) tmpa++;
    if(tmpa % 2 == 0) {
        tmpa = 1 << tmpa / 2;
        for(int i = 0; i < n; i++)
            for(int j = 0; j < n; j++)
                if(((i / tmpa % 2) ^ (j / tmpa % 2))) A[i][j] = 0;
    } else {
        tmpa = 1 << tmpa  / 2;
        for(int i = 0; i < n; i++)
            for(int j = 0; j < n; j++)
                if((i / tmpa % 2)) A[i][j] = 0;
    }
    int tmpb = 0;
    while(b % (1 << (tmpb + 1)) == 0) tmpb++;
    if(tmpb % 2 == 0) {
        tmpb = 1 << tmpb / 2;
        for(int i = 0; i < n; i++)
            for(int j = 0; j < n; j++)
                if(((i / tmpb % 2) ^ (j / tmpb % 2))) A[j][i] = 0;
    } else {
        tmpb = 1 << tmpb / 2;
        for(int i = 0; i < n; i++)
            for(int j = 0; j < n; j++)
                if((i / tmpb % 2)) A[j][i] = 0;
    }
    int cnt = 0;
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++)
            if(A[i][j] && ++cnt <= n * n / 4) printf("%d %d\n", i, j);
    return 0;
}