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 mset(a, b) memset(a, b, sizeof(a))
#define mcpy(a, b) memcpy(a, b, sizeof(a))
#define rg register
using namespace std;
typedef long long LL;
const int N = 15;
const LL MOD = 1e9 + 7;

template <typename T> inline void read(T &AKNOI) {
	T x = 0, flag = 1;
	char ch = getchar();
	while (!isdigit(ch)){
		if (ch == '-') flag = -1;
		ch = getchar();
	}
	while (isdigit(ch)) {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	AKNOI = flag * x;
}

int n, m, g[N], bit[1<<N];
LL pw[N*N], dp[1<<N];

void init() {
	read(n); read(m);
	pw[0] = 1;
	for (int i = 1, u, v; i <= m; ++i) {
		pw[i] = pw[i-1] * 2 % MOD;
		read(u); read(v);
		--u; --v;
		g[u] |= (1<<v);
	}
}

void solve() {
	int full = (1<<n);
	dp[0] = 1;
	for (int S = 1; S < full; ++S) {
		bit[S] = bit[S>>1] + (S&1);
	}
	for (int S = 0; S < full; ++S) {
		if ((S & 1) ^ ((S>>1) & 1)) continue;
		for (int T = S; T; T = (T-1) & S) {
			if ((T & 1) ^ ((T>>1) & 1)) continue;
			LL tmp = 1;
			for (int i = 0; i < n; ++i) {
				if ((S>>i) & 1) {
					if ((T>>i) & 1) {
						tmp = tmp * pw[bit[ g[i] & (S ^ T) ]] % MOD;
					} else {
						tmp = tmp * (pw[bit[ g[i] & T ]] - 1) % MOD;
					}
				}
			}
			dp[S] = (dp[S] + tmp % MOD * dp[S^T]) % MOD;
		}
	}
	printf("%lld\n", (pw[m] - dp[full-1] + MOD) % MOD);
}

int main() {
	init();
	solve();
	return 0;
}