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 <algorithm>
#include <iostream>
#include <string>
#include <vector>
using namespace std;

typedef long long i64;
const int MAXN = 1e5 + 10;
const int MOD = 1e9 + 7;

int answer = 1;
int N, M;
int edgeCnt[20][MAXN], dp[MAXN];

inline int loBit(int x)
{
	return x & (-x);
}

inline bool checkContain(int x)
{
	return !(((x & 0x03) != 0) && ((x & 0x03) != 0x03));
}

int main()
{
	cin >> N >> M ;

	for(int i = 1; i <= M; i++)
	{
		int u, v;
		cin >> u >> v, u--, v--;
		edgeCnt[u][1 << v] += 1;
		answer = (2LL * answer) % MOD;
	}

	for(int i = 0; i < N; i++)
		for(int j = 0; j < (1 << N); j++)
			edgeCnt[i][j] = edgeCnt[i][loBit(j)] + edgeCnt[i][j ^ loBit(j)];

	for(int i = 0; i < (1 << N); i++)
	{
		if(!checkContain(i))
			continue;
		dp[i] = 1;

		for(int j = ((i - 1) & i); j > 0; j = ((j - 1) & i))
		{
			if(!checkContain(j))
				continue;

			int tmp = i ^ j;
			i64 val = dp[tmp];

			for(int k = 0; k < N; k++)
			{
				if(tmp & (1 << k))
					val = 1LL * val * ((1 << edgeCnt[k][j]) - 1) % MOD;
				else if(j & (1 << k))
					val = 1LL * val * (1 << edgeCnt[k][tmp]) % MOD;
			}
			dp[i] = (1LL * dp[i] + val) % MOD;
		}
	}
	cout << (answer - dp[(1<<N) - 1] + MOD) % MOD << endl ;
	return 0;
}