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;
typedef long long ll;
const int N = 20, mod = 1e9 + 7;
int n, m, cnt, e[N], f[1 << N], sz[1 << N], ans;

int main() {
	scanf("%d%d", &n, &m), cnt = (1 << n) - 1, f[0] = 1;
	for(int i = 1, x, y; i <= m; i++) scanf("%d%d", &x, &y), --x, --y, e[x] |= (1 << y);
	for(int i = 1; i <= cnt; i++) sz[i] = sz[i >> 1] + (i & 1);
	for(int i = 0; i <= cnt; i++) sz[i] = 1 << sz[i];
	for(int i = 2; i <= cnt; i++) if((i & 1) == ((i >> 1) & 1)) {
		for(int j = i; j; j = i & (j - 1)) if((j & 1) == ((j >> 1) & 1)) {
			int w = 1, v = (i ^ j);
			for(int k = 0; k < n; k++) if(i & (1 << k)) {
				if(j & (1 << k)) w = (ll)w * sz[e[k] & v] % mod;
				else w = (ll)w * (sz[e[k] & j] - 1) % mod;
			}
			f[i] = (f[i] + (ll)w * f[v] % mod) % mod;
		}
	}
	ans = 1; for(int i = 1; i <= m; i++) ans = (ans + ans) % mod;
	printf("%d\n", (ans + mod - f[cnt]) % mod);
	return 0;
}