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
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;

typedef long long s64;

const int MaxN = 15;
const int MaxM = MaxN * (MaxN - 1) / 2;
const int M = 1000000007;

int n, m;
int out[MaxN];

int mul[1 << MaxN];

int f[1 << MaxN];

int main()
{
	cin >> n >> m;
	for (int i = 0; i < m; ++i)
	{
		int u, v;
		cin >> u >> v, --u, --v;

		out[u] |= 1 << v;
	}

	mul[0] = 1;
	for (int s = 1; s < 1 << n; ++s)
		mul[s] = mul[s ^ (s & -s)] << 1;

	f[0] = 1;
	for (int s = 1; s < 1 << n; ++s)
		if ((s ^ (~s >> 1)) & 1)
		{
			for (int t = s; t; --t &= s)
			{
				int v = s ^ t;

				int sum = f[v];
				for (int i = 0; i < n; ++i)
				{
					if (v >> i & 1)
						sum = (s64)sum * (mul[out[i] & t] - 1) % M;
					if (t >> i & 1)
						sum = (s64)sum * mul[out[i] & v] % M;
				}

				f[s] = (f[s] + sum) % M;
			}
		}

	int res = 1;
	for (int i = 0; i < m; ++i)
		res = res * 2 % M;

	cout << (res + M - f[(1 << n) - 1]) % M << endl;

	return 0;
}