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;

const int N = 610;

int n, col[N][N], cnt[3];
vector<pair<int,int>> diff;

void dfs(int x, int y, int c) {
    if (x >= 0 && x < 2*n && y >= 0 && y < 2*n && col[x][y] == 0) {
        col[x][y] = c;
        cnt[c]++;
        for (auto d : diff) dfs(x+d.first, y+d.second, 3-c);
    }
}

int main() {
    cin >> n;

    int iter = 2; while (iter--) {
        // calcular arcos
        int d; cin >> d;
        diff.clear();
        for (int dx = 0; dx < 2*n; dx++) 
            for (int dy = 0; dy < 2*n; dy++) 
                if (dx*dx + dy*dy == d) {
                    diff.emplace_back(dx,dy);
                    diff.emplace_back(-dx,dy);
                    diff.emplace_back(dx,-dy);
                    diff.emplace_back(-dx,-dy);
                }

        // partir bipartito
        for (int i = 0; i < 2*n; i++) 
            for (int j = 0; j < 2*n; j++) if (col[i][j] == 0) 
                dfs(i,j,1);

        // quedarse con particion mayor
        int best = (cnt[1] > cnt[2] ? 1 : 2);
        for (int i = 0; i < 2*n; i++) 
            for (int j = 0; j < 2*n; j++) 
                col[i][j] = (col[i][j] == best ? 0 : 3);
    }

    int need = n*n;
    for (int i = 0; i < 2*n; i++) 
        for (int j = 0; j < 2*n; j++) if (col[i][j] == 0) {
            cout << i << ' ' << j << endl;
            if (--need == 0) return 0;
        }

    return 0;
}
