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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define MOD 1000000007
#define ADD(X,Y) ((X) = ((X) + (Y)%MOD) % MOD)
typedef long long i64; typedef vector<int> ivec; typedef vector<string> svec;

int N, M;
int X[200], Y[200];
int out[200];
i64 dp[1 << 15];

int main()
{
	scanf("%d%d", &N, &M);
	for (int i = 0; i < M; ++i) {
		scanf("%d%d", X + i, Y + i);
		--X[i]; --Y[i];
		out[X[i]] |= 1 << Y[i];
	}
	dp[0] = 1;
	for (int m = 1; m < (1 << N); ++m) {
		dp[m] = 0;
		for (int t = m; t > 0; t = (t - 1) & m) {
			// choose <t> for nodes with largest grundy number among <m>
			if ((t & 3) == 3) continue;
			int rem = m ^ t;
			i64 pat = dp[rem];

			for (int i = 0; i < N; ++i) {
				if (!(m & (1 << i))) {
					pat = pat * ((1 << __builtin_popcount(out[i] & t)) - 1) % MOD;
				}
				if (t & (1 << i)) {
					pat = pat * (1 << __builtin_popcount(out[i] & ~m)) % MOD;
				}
			}
			ADD(dp[m], pat);
		}
	}
	printf("%lld\n", dp[(1 << N) - 1]);
	return 0;
}
