Problem: There is a directed graph G with N vertices and M edges. The vertices are numbered 1 through N, and the edges are numbered 1 through M. Edge i is directed from x_i to y_i. Here, x_i < y_i holds. Also, there are no multiple edges in G.

Consider selecting a subset of the set of the M edges in G, and removing these edges from G to obtain another graph G'. There are 2^M different possible graphs as G'.

Alice and Bob play against each other in the following game played on G'. First, place two pieces on vertices 1 and 2, one on each. Then, starting from Alice, Alice and Bob alternately perform the following operation:

* Select an edge i such that there is a piece placed on vertex x_i, and move the piece to vertex y_i (if there are two pieces on vertex x_i, only move one). The two pieces are allowed to be placed on the same vertex.



The player loses when he/she becomes unable to perform the operation. We assume that both players play optimally.

Among the 2^M different possible graphs as G', how many lead to Alice's victory? Find the count modulo 10^9+7.

Constraints

* 2 ≤ N ≤ 15
* 1 ≤ M ≤ N(N-1)/2
* 1 ≤ x_i < y_i ≤ N
* All (x_i,\ y_i) are distinct.

Input

Input is given from Standard Input in the following format:


N M
x_1 y_1
x_2 y_2
:
x_M y_M


Output

Print the number of G' that lead to Alice's victory, modulo 10^9+7.

Examples

Input

2 1
1 2


Output

1


Input

3 3
1 2
1 3
2 3


Output

6


Input

4 2
1 3
2 4


Output

2


Input

5 10
2 4
3 4
2 5
2 3
1 2
3 5
1 3
1 5
4 5
1 4


Output

816

Solution (in CPP):

Solution:
#include <cstdio>
#define MOD 1000000007
typedef long long ll;
int cnt[25][100005], dp[100005];
int main()
{
	// freopen("AGC016-F.in", "r", stdin);
	int n, m, ans = 1;
	scanf("%d%d", &n, &m);
	for (int i = 0; i < m; i++)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		cnt[--u][1 << --v]++;
		(ans += ans) %= MOD;
	}
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < 1 << n; j++)
			cnt[i][j] = cnt[i][j & -j] + cnt[i][j ^ (j & -j)];
	}
	for (int i = 0; i < 1 << n; i++)
	{
		if ((i & 3) && (i & 3) != 3)
			continue;
		dp[i] = 1;
		for (int j = i - 1 & i; j; j = j - 1 & i)
		{
			if ((j & 3) && (j & 3) != 3)
				continue;
			int x = i ^ j, val = dp[x];
			for (int k = 0; k < n; k++)
			{
				if (x & (1 << k))
					val = (ll)val * ((1 << cnt[k][j]) - 1) % MOD;
				else if (j & (1 << k))
					val = (ll)val * (1 << cnt[k][x]) % MOD;
			}
			(dp[i] += val) %= MOD;
		}
	}
	printf("%d\n", (ans - dp[(1 << n) - 1] + MOD) % MOD);
	return 0;
}