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 <bits/stdc++.h>
using namespace std;

#define debug(...) fprintf(stderr, __VA_ARGS__)

const int mod = 1e9 + 7;

int n, m;
int two[300 + 5], adj[15 + 5], dp[(1 << 15) + 5];

int main() {
	scanf("%d%d", &n, &m);
	two[0] = 1;
	for (int i = 0; i < m; ++i) two[i + 1] = (two[i] << 1) % mod;
//	for (int i = 0; i <= m; ++i) debug("%d %d\n", i, two[i]);
	for (int i = 0; i < m; ++i) {
		int x, y;
		scanf("%d%d", &x, &y), --x, --y;
		adj[x] |= 1 << y;
//		adj[y] |= 1 << x;
	}
//	for (int i = 0; i < n; ++i) debug("%d %d\n", i, adj[i]);
	dp[0] = 1;
	for (int S = 1; S < (1 << n); ++S) {
		if ((S & 1) != (S >> 1 & 1)) continue;
		for (int T = S; T; (--T) &= S) {
			if ((T & 1) != (T >> 1 & 1)) continue;
//			debug("%d %d\n", S, T);
			int ways = 1, U = S ^ T;
			for (int i = 0; i < n; ++i)
				if (S >> i & 1) {
					if (T >> i & 1) ways = (long long)ways * two[__builtin_popcount(adj[i] & U)] % mod;
					else ways = (long long)ways * (two[__builtin_popcount(adj[i] & T)] - 1) % mod;
				}
			(dp[S] += (long long)dp[S ^ T] * ways % mod) %= mod;
		}
	}
//	for (int i = 0; i < (1 << n); ++i) debug("%d\n", dp[i]);
	printf("%d\n", (two[m] - dp[(1 << n) - 1] + mod) % mod);
}