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 ctz __builtin_ctz
#define popc __builtin_popcount

typedef long long LL;
const int N = 15, mod = 1000000007;

int n, m, dp[1 << N], e[N], mask;

void reduce(int &x) { x += x >> 31 & mod; }
int pow(int x, int y, int ans = 1) {
	for (; y; y >>= 1, x = (LL) x * x % mod)
		if (y & 1) ans = (LL) ans * x % mod;
	return ans;
}
bool valid(int s) { return (s & 1) == (s >> 1 & 1); }

int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::cin >> n >> m, mask = ~(-1 << n);
	for (int i = 0, x, y; i < m; ++i)
		std::cin >> x >> y, --x, --y, e[x] |= 1 << y;
	dp[0] = 1;
	for (int i = 0; i <= mask; ++i) if (dp[i]) {
		int c = mask ^ i;
		for (int j = c; j; j = j - 1 & c) if (valid(j)) {
			int prod = dp[i];
			for (int _k = i, k = ctz(_k); _k; k = ctz(_k &= _k - 1))
				prod = (LL) prod * ~(-1 << popc(e[k] & j)) % mod;
			for (int _k = j, k = ctz(_k); _k; k = ctz(_k &= _k - 1))
				prod = (LL) prod * (1 << popc(e[k] & i)) % mod;
			reduce(dp[i | j] += prod - mod);
		}
	}
	std::cout << (pow(2, m) - dp[mask] + mod) % mod << '\n';
	return 0;
}