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 ll long long
#define si(x) scanf("%d",&x)
#define sl(x) scanf("%lld",&x)
#define pi(x) printf("%d",x)
#define pl(x) printf("%lld",x)
#define pb push_back
#define mkp make_pair
#define fi first
#define se second
#define re register
#define rep(i,m,n) for(int i=m;i<=n;i++)
#define per(i,n,m) for(int i=m;i>=n;i--)
#define rrep(i,m,n) for(register int i=m;i<=n;i++)
#define rper(i,n,m) for(register int i=m;i>=n;i--)
#define ull unsigned  long long
using namespace std;
const int INF =1e9+10;
const int N = 1e3 + 10;
int n,d1,d2,cnt;
int a[N][N];
void solve(int x){
    int tmp=0;
    while(x%4==0){
        x/=4;
        tmp++;
    }
    if(x&1){
        rep(i,0,n*2-1){
            rep(j,0,n*2-1){
                if(((i>>tmp)+(j>>tmp))&1){
                    a[i][j]=1;
                }
            }
        }
    }else{
        rep(i,0,n*2-1){
            rep(j,0,n*2-1){
                if(((i>>tmp))&1){
                    a[i][j]=1;
                }
            }
        }
    }
}
signed main() {
    si(n);
    si(d1);
    si(d2);
    solve(d1);
    solve(d2);
    rep(i,0,n*2-1){
        rep(j,0,n*2-1){
            if(cnt<n*n&&!a[i][j]){
                printf("%d %d\n",i,j);
                cnt++;
            }
        }
    }
    return 0;
}
