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 <cstdio>
#include <vector>
const long long MOD = 1000000007;
int N, M, sg[16], c0[15], c1[15], e[16][16];
std::vector < int > V[16];
long long O = 0;
long long DFS(int k)
{
	if (k <= 2)
	{
		static long long H[15];
		static int T[15];
		H[0] = 1;
		for (int i = 1; H[i - 1] && i <= N - 1; i++)
			H[i] = H[i - 1] * ((1 << c0[i - 1]) - 1) * ((1 << c1[i - 1]) - 1) % MOD;
		T[N - 1] = 0;
		for (int i = N - 2; ~i; i--)
			T[i] = T[i + 1] + c0[i + 1] + c1[i + 1];
		long long O = 0;
		for (int i = 0; i <= N - 1 && H[i]; i++)
			O = (O + (H[i] << T[i])) % MOD;
		return O;
	}
	static int cnt[16];
	for (int i = 0; i <= N; i++)
		cnt[i] = 0;
	for (int i : V[k])
		cnt[sg[i]]++;
	static long long H[16], T[16];
	H[0] = 1;
	for (int i = 1; i <= N; i++)
		H[i] = H[i - 1] * ((1 << cnt[i - 1]) - 1) % MOD;
	T[N] = 1;
	for (int i = N - 1; ~i; i--)
		T[i] = (T[i + 1] << cnt[i + 1]) % MOD;
	long long U[16], O = 0;
	for (int i = 0; i <= N; i++)
		U[i] = H[i] * T[i] % MOD;
	for (int i = 0; i <= N; i++)
		if (U[i])
		{
			if (e[1][k])
				c0[i]++;
			if (e[2][k])
				c1[i]++;
			sg[k] = i;
			O = (O + U[i] * DFS(k - 1)) % MOD;
			if (e[1][k])
				c0[i]--;
			if (e[2][k])
				c1[i]--;
		}
	return O;
}
int main()
{
	scanf("%d%d", &N, &M);
	long long BASE = 1;
	for (int i = 1, u, v; i <= M; i++)
	{
		scanf("%d%d", &u, &v);
		BASE = BASE * 2 % MOD;
		if (u != 1 || v != 2)
		{
			V[u].push_back(v);
			e[u][v] = 1;
		}
	}
	printf("%lld\n", (BASE - DFS(N) + MOD) % MOD);
}