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
//  Created by Sengxian on 2017/6/30.
//  Copyright (c) 2017年 Sengxian. All rights reserved.
//  AGC016F 博弈 + 状压 DP
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
inline int readInt() {
	int n = 0, ch = getchar();
	while (!isdigit(ch)) ch = getchar();
	while (isdigit(ch)) n = n * 10 + ch - '0', ch = getchar();
	return n;
}

const int MAX_N = 15, MOD = 1000000007;
int n, m, G[MAX_N], popCount[1 << MAX_N], cnt[MAX_N][1 << MAX_N];
int dp[1 << MAX_N];

inline bool check(int s) {
	return (s & 3) == 0 || (s & 3) == 3;
}

void prepare() {
	popCount[0] = 0;
	for (int s = 1; s < (1 << n); ++s)
		popCount[s] = popCount[s >> 1] + (s & 1);
	for (int s = 0; s < (1 << n); ++s)
		for (int i = 0, t = s; i < n; ++i, t &= ~(1 << i))
			cnt[i][s] = popCount[t & G[i]];
}

void solve() {
	dp[0] = 1;
	for (int s = 1; s < (1 << n); ++s) if (check(s)) {
		int t = s;
		for (int u = s; u; u = (u - 1) & s) {
			int t = s ^ u, res = dp[t];
			if (!check(t)) continue;
			// u all zero
			// t all non-zero
			// u -> t
			for (int i = 0; i < n; ++i) if (u >> i & 1)
				res = (ll)res * (1 << cnt[i][t]) % MOD;
			// t -> u
			for (int i = 0; i < n; ++i) if (t >> i & 1)
				res = (ll)res * ((1 << cnt[i][u]) - 1) % MOD;
			(dp[s] += res) %= MOD;	
		}
	}
	
	int res = 1;
	for (int i = 0; i < m; ++i) (res *= 2) %= MOD;
	printf("%d\n", (res - dp[(1 << n) - 1] + MOD) % MOD);
}

int main() {
	n = readInt(), m = readInt();
	for (int i = 0; i < m; ++i) {
		int u = readInt() - 1, v = readInt() - 1;
		G[u] |= 1 << v;
	}
	
	prepare();
	solve();
	
	return 0;
}
