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 = 310;

typedef pair<int, int> P;
typedef long long LL;

int G[2][N * 2][N * 2], n;

vector<P> dir;

void dfs(int x, int y, int t) {
    for (auto v : dir) {
        int tx = x + v.first, ty = y + v.second;
        if (tx > n * 2 || ty > n * 2 || tx < 1 || ty < 1 || G[t][tx][ty]) continue;
        G[t][tx][ty] = 3 - G[t][x][y], dfs(tx, ty, t);
    }
}

void Just_DOIT(int t, int d) {
    dir.clear();
    for (int i = 0; i < 2 * n; i++)
        for (int j = 0; j < 2 * n; j++)
            if ((LL)i * i + (LL)j * j == d) 
                for (int t = 0; t < 4; t++) 
                    dir.push_back(P(t & 1 ? i : -i, t & 2 ? j : -j));
    for (int i = 1; i <= n * 2; i++)
        for (int j = 1; j <= n * 2; j++) if (!G[t][i][j]) 
            G[t][i][j] = 1, dfs(i, j, t);
}

vector<P> fuck[4];

int main() {
    int d1, d2; scanf("%d%d%d", &n, &d1, &d2);
    Just_DOIT(0, d1), Just_DOIT(1, d2);
    for (int i = 1; i <= n * 2; i++)
        for (int j = 1; j <= n * 2; j++)
            fuck[G[0][i][j] + G[1][i][j] * 2 - 3].push_back(P(i, j));
    for (int i = 0; i < 4; i++) if (fuck[i].size() >= n * n) {
        for (int j = 0; j < n * n; j++) printf("%d %d\n", fuck[i][j].first - 1, fuck[i][j].second - 1);
        return 0;
    }
}
