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
def check_odd(i, j):
    return (i + j) % 2 == 0


def check_even(i, j):
    return i % 2 == 0


def solve(n, d1, d2):
    s1, s2 = 0, 0
    while d1 % 4 == 0:
        d1 >>= 2
        s1 += 1
    while d2 % 4 == 0:
        d2 >>= 2
        s2 += 1

    f1 = check_odd if d1 % 2 else check_even
    f2 = check_odd if d2 % 2 else check_even
    lim = n ** 2

    buf = []
    cnt = 0
    for i in range(2 * n):
        _i1 = i >> s1
        _i2 = i >> s2
        for j in range(2 * n):
            if f1(_i1, j >> s1) and f2(_i2, j >> s2):
                buf.append('{} {}'.format(i, j))
                cnt += 1
                if cnt == lim:
                    return buf


n, d1, d2 = map(int, input().split())
print('\n'.join(solve(n, d1, d2)))
