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>
#define x first
#define y second
using namespace std;

typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
const ll MAX=605;
const ll INF=0x3f3f3f3f;
const ll MOD=998244353;
int n,d1,d2,s1=1,s2=1,mask1[MAX][MAX],mask2[MAX][MAX],cnt;
int main(){
    scanf("%d%d%d",&n,&d1,&d2);
    while(d1%4==0) d1/=4,s1*=2;
    while(d2%4==0) d2/=4,s2*=2;
    for(int i=0;i<2*n;i++) for(int j=0;j<2*n;j++){
        if(d1%2){if((i/s1)%2==(j/s1)%2) mask1[i][j]=1;}
        else{if((i/s1)%2==0) mask1[i][j]=1;}
    }
    for(int i=0;i<2*n;i++) for(int j=0;j<2*n;j++){
        if(d2%2){if((i/s2)%2==(j/s2)%2) mask2[i][j]=1;}
        else{if((i/s2)%2==0) mask2[i][j]=1;}
    }
    for(int i=0;i<2*n;i++) for(int j=0;j<2*n;j++) if(mask1[i][j]&&mask2[i][j]){
        cnt++;
        printf("%d %d\n",i,j);
        if(cnt==n*n) return 0;
    }
}
