Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define popc __builtin_popcount
#define ctz __builtin_ctz

typedef long long ll;
const int N = 18, N2 = 32854, mod = 1000000007;

int V, E, ALL;
int G[N], f[N2];

int main() {
	int i, j, u, v, S, tot, win, lose, ans = 1; ll cur;
	scanf("%d%d", &V, &E), ALL = ~(-1 << V);
	for (i = 0; i < E; ++i)
		scanf("%d%d", &u, &v), G[--u] |= 1 << --v, ans = ans * 2 % mod;
	for (*f = 1, i = 2; i <= ALL; i += 2)
		for (j = i; ; j = (j - 1) & i) {
			cur = 1, tot = i + (i >> 1 & 1);
			win = j + (j >> 1 & 1), lose = tot & ~win;
			for (S = i + (i >> 1 & 1); S; S &= S - 1)
				v = ctz(S), cur = cur * (win >> v & 1 ? (1 << popc(G[v] & lose)) - 1 : 1 << popc(G[v] & win)) % mod;
			f[i] = (f[i] + cur * f[j]) % mod;
			if (!j) break;
		}
	ans -= f[ALL - 1], printf("%d\n", ans + (ans >> 31 & mod));
	return 0;
}