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<bits/stdc++.h>
using namespace std;


int n;
int bd[601][601];

void do_ban(int x){
    int m = 1;
    while(x % 4 == 0){
        m *= 2;
        x /= 4;
    }
    if(x % 4 == 3) return;
    if(x % 4 == 1){
        for(int i=0;i<2*n;i++){
            for(int j=0;j<2*n;j++){
                if((i % (2 * m) >= m) ^ (j % (2 * m) >= m)){
                    bd[i][j] = 0;
                }
            }
        }
    }else if(x % 4 == 2){
        for(int i=0;i<2*n;i++){
            for(int j=0;j<2*n;j++){
                if(i % (2 * m) >= m){
                    bd[i][j] = 0;
                }
            }
        }
    }
}

int main(){
    int d1, d2;
    scanf("%d%d%d",&n,&d1,&d2);
    for(int i=0;i<2*n;i++){
        for(int j=0;j<2*n;j++){
            bd[i][j] = 1;
        }
    }
    do_ban(d1);
    do_ban(d2);
    int cnt = 0;
    for(int i=0;i<2*n;i++){
        for(int j=0;j<2*n;j++){
            if(cnt == n*n) continue;
            if(bd[i][j]){
                printf("%d %d\n", i, j);
                cnt++;
            }
        }
    }
}
