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 color[600][600][2];
int N, D[2];
vector<pair<int, int>> pr[2];

void dfs(int i, int j, int k, int c){
    if(color[i][j][k]) return;
    color[i][j][k] = c;

    for(auto& p : pr[k]){
        int x = i+p.first, y = j+p.second;
        if(0<=x && x<2*N && 0<=y && y<2*N) dfs(x, y, k, -c);
    }
}

int main(){
    cin >> N >> D[0] >> D[1];
    
    for(int i=-2*N; i<=2*N; i++) for(int j=-2*N; j<=2*N; j++){
        for(int k=0; k<2; k++) if(i*i + j*j == D[k]) pr[k].push_back({i, j});
    }

    for(int i=0; i<2*N; i++) for(int j=0; j<2*N; j++) for(int k=0; k<2; k++) dfs(i, j, k, 1);

    map<int, vector<pair<int, int>>> mp;
    for(int i=0; i<2*N; i++) for(int j=0; j<2*N; j++){
        int code = 10*color[i][j][0] + color[i][j][1];
        mp[code].push_back({i, j});
    }
    for(auto& p : mp){
        auto v = p.second;
        if(v.size() >= N*N){
            int num = 0;
            for(auto& a : v){
                printf("%d %d\n", a.first, a.second);
                num++;
                if(num == N*N) return 0;
            }
        }
    }
    return 0;
}