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>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define ll long long
#define mod 1000000007
#define N 16
#define lowbit(x) ((x) & -(x))
int f[1 << N], g[N][1 << N], mul[N * N];
int n, m;
int main()
{
	scanf("%d%d", &n, &m);
	mul[0] = 1;
	for (int i = 1; i <= m; i++)
		mul[i] = mul[i - 1] * 2ll % mod;
	for (int i = 1; i <= m; i++)
	{
		int x, y;
		scanf("%d%d", &x, &y);
		g[x - 1][1 << (y - 1)]++;
	}
	for (int i = 0; i < n; i++)
		for (int j = 1; j < (1 << n); j++)
			g[i][j] = g[i][j - lowbit(j)] + g[i][lowbit(j)];
	for (int s = 1; s < (1 << n); s++)
	{
		if ((s & 1) && !(s & 2))continue;
		if (!(s & 1) && (s & 2))continue;
		f[s] = 1;
		for (int t = s; t; t = (t - 1) & s)
		{
			if ((t & 1) && !(t & 2))continue;
			if (!(t & 1) && (t & 2))continue;
			int u = s - t;
			int w = f[u];
			for (int i = 0; i < n; i++)
			{
				if (u & (1 << i))
					w = (ll)(mul[g[i][t]] - 1) * w % mod;
				if (t & (1 << i))
					w = (ll)mul[g[i][u]] * w % mod;
			}
			f[s] = (f[s] + w) % mod;
		}
	}
	int ans = (mul[m] - f[(1 << n) - 1]) % mod;
	if (ans < 0)
		ans += mod;
	printf("%d\n", ans);
}