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>
using namespace std;

const int N = 15, MOD = 1000 * 1000 * 1000 + 7;

int sum(int a, int b) {
	a += b;
	if (a < 0)
		a += MOD;
	else if (a >= MOD)
		a -= MOD;
	return a;
}
void _sum(int &a, int b) {
	a = sum(a, b);
}
int mul(int a, int b) {
	return 1LL * a * b % MOD;
}
void _mul(int &a, int b) {
	a = mul(a, b);
}

int count(int x) {
	return __builtin_popcount(x);
}

int n, m, nei[N], dp[1 << N];
int ans = 1;

void input() {
	cin >> n >> m;
	while (m--) {
		int x, y;
		cin >> x >> y;
		nei[--x] |= 1 << (--y);
		_mul(ans, 2);
	}
}

int f(int big, int small) {
	int res = 1;
	for (int i = 0; i < n; i++)
		if (big >> i & 1)
			_mul(res, (1 << count(nei[i] & small)) - 1);
		else if (small >> i & 1)
			_mul(res, 1 << count(nei[i] & big));
	return res;
}

bool ok(int mask) {
	return (mask & 1) == (mask >> 1 & 1);
}

void calc() {
	dp[0] = 1;
	for (int mask = 1; mask < (1 << n); mask++)
		if (ok(mask))
			for (int sub = mask; sub; sub = --sub & mask)
				if (ok(sub) && ok(mask ^ sub))
					_sum(dp[mask], mul(dp[mask ^ sub], f(mask ^ sub, sub)));
}

int main() {
	ios::sync_with_stdio(false), cin.tie(0);
	input();
	calc();
	cout << sum(ans, -dp[(1 << n) - 1]);
}