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 <algorithm>
using namespace std;

const int MOD = int(1E9) + 7;

inline int add(int x, int y) {return (x + y >= MOD ? x + y - MOD : x + y);}
inline int sub(int x, int y) {return (x - y < 0 ? x - y + MOD : x - y);}
inline int mul(int x, int y) {return 1LL * x * y % MOD;}

bool check(int s) {return (s & 1) + ((s >> 1) & 1) == 2;}
int lowbit(int x) {return (x & -x);}

int f[1 << 15], g[1 << 15], bts[1 << 15], lg[1 << 15], pw2[20];

int G[15], N, M, S;
int main() {
	scanf("%d%d", &N, &M), S = (1 << N);
	for(int i=1;i<=M;i++) {
		int x, y; scanf("%d%d", &x, &y), x--, y--;
		G[x] |= (1 << y);
	}
	f[0] = pw2[0] = 1;
	for(int i=0;i<N;i++) lg[1 << i] = i;
	for(int i=1;i<=N;i++) pw2[i] = mul(2, pw2[i-1]);
	for(int s=1;s<S;s++) bts[s] = bts[s >> 1] + (s & 1);
	for(int s=0;s<S;s++) {
		for(int s2=s;s2;s2=s&(s2-1)) {
			if( check(s2) ) continue;
			int s3 = (s ^ s2), p = s2, del = f[s3];
			while( p ) {
				int q = lowbit(p);
				del = mul(del, pw2[bts[G[lg[q]] & s3]]);
				p ^= q;
			}
			p = s3;
			while( p ) {
				int q = lowbit(p);
				del = mul(del, sub(pw2[bts[G[lg[q]] & s2]], 1));
				p ^= q;
			}
			f[s] = add(f[s], del);
		}
	}
	printf("%d\n", f[S - 1]);
}