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>
const int MOD = int(1E9) + 7;
const int MAXN = 15;
int nxt[MAXN], dp[1<<MAXN];
int pw[MAXN*MAXN], f[1<<MAXN];
int lowbit(int x) {
	return x & -x;
}
int main() {
	int N, M;
	scanf("%d%d", &N, &M);
	for(int i=1;i<=M;i++) {
		int x, y;
		scanf("%d%d", &x, &y);
		x--, y--;
		nxt[x] |= (1<<y);
	}
	pw[0] = 1;
	for(int i=1;i<=MAXN*MAXN;i++)
		pw[i] = 1LL * pw[i-1] * 2 % MOD;
	int ans = pw[M], t = (1<<N) - 1;
	for(int i=1;i<=t;i++)
		f[i] = f[i^lowbit(i)] + 1;
	dp[0] = 1;
	for(int s=1;s<=t;s++) {
		for(int s1=s;;s1=s&(s1-1)) {
			if( (!(s1&1)) != (!(s1&2)) ) continue;
			int s2 = s ^ s1, del = dp[s1];
			for(int i=0;i<N;i++) {
				if( (1<<i) & s ) {
					if( (1<<i) & s1 ) {
						int p = (nxt[i] & s2);
						del = 1LL * del * (pw[f[p]] - 1) % MOD;
					}
					else {
						int p = (nxt[i] & s1);
						del = 1LL * del * pw[f[p]] % MOD;
					}
				}
			}
			dp[s] = (dp[s] + del) % MOD;
			if( !s1 ) break;
		}
	}
	printf("%d\n", ((ans - dp[t]) % MOD + MOD) % MOD);
}