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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#define N 17
#define NN 34567
#define NNN 15671000
typedef long long ll;
template<typename T> inline void read(T &x) {
	x = 0; char c = getchar();
	while (!isdigit(c)) c = getchar();
	while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
}
using namespace std;
const int P = 1e9 + 7;
ll mi[N * N];
int n, m, All;
int e[N][N];
int ecnt[N][NN];//ecnt[i][s] : i -> s
inline void init() {
	mi[0] = 1;
	for (int i = 1; i <= m; ++i)	mi[i] = (mi[i - 1] << 1) % P;
	for (int i = 0; i < n; ++i) {
		for (int s = 0; s <= All; ++s) {
			for (int j = 0; j < n; ++j) if ((s >> j) & 1) {
				ecnt[i][s] += e[i][j];
			}
		}
	}
}
inline ll calc1(int s, int t) {
	ll res = 1;
	for (int i = 0; i < n; ++i) if ((s >> i) & 1) {
		res = 1ll * res * mi[ecnt[i][t]] % P;
	}
	return res;
}
inline ll calc2(int s, int t) {
	ll res = 1;
	for (int i = 0; i < n; ++i) if ((s >> i) & 1) {
		res = 1ll * res * (mi[ecnt[i][t]] - 1) % P;
	}
	return res;
}
int f[NNN];
inline void work() {
	f[0] = 1;
	for (int s = 1; s <= All; ++s) if ((s & 1) == ((s >> 1) & 1)) {
		for (int t = s; t; t = (t - 1) & s) {
			f[s] = (f[s] + 1ll * calc1(t, s ^ t) * calc2(s ^ t, t) % P * f[s ^ t] % P) % P;
		}
	}
	ll ans = (mi[m] - f[All]) % P;
	printf("%lld\n", (ans % P + P) % P);
}
int main() {
	read(n), read(m); All = (1 << n) - 1;
	for (int i = 1; i <= m; ++i) {
		int u, v; read(u), read(v);
		--u; --v;
		e[u][v] = 1;
	}
	init();
	work();
	return 0;
}
