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
#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll mod = 1000000007;
ll dp[1 << 15];
vector<int>pat[15];
int cnt[15][1 << 15];
int main()
{
	int num, way;
	scanf("%d%d", &num, &way);
	for (int i = 0; i < way; i++)
	{
		int za, zb;
		scanf("%d%d", &za, &zb);
		za--, zb--;
		pat[za].push_back(zb);
	}
	for (int p = 0; p < (1 << num); p++)
	{
		for (int i = 0; i < num; i++)
		{
			for (int j = 0; j < pat[i].size(); j++)
			{
				if (p&(1 << pat[i][j]))cnt[i][p]++;
			}
		}
	}
	dp[0] = 1;
	for (int p = 0; p < (1 << num); p++)
	{
		if (((p & 1) == 0) != ((p & 2) == 0))
		{
			dp[p] = 0;
			continue;
		}
		//printf("%d %lld\n", p, dp[p]);
		int v[15];
		int pt = 0;
		for (int i = 0; i < num; i++)if ((p&(1 << i)) == 0)v[pt++] = i;
		for (int q = 0; q < (1 << pt); q++)
		{
			int x = 0;
			for (int i = 0; i < pt; i++)if (q&(1 << i))x += 1 << v[i];
			if (x == 0)continue;
			ll tim = 1;
			for (int i = 0; i < pt; i++)if ((q&(1 << i)) == 0)tim = (tim*((1 << cnt[v[i]][x]) - 1)) % mod;
			for (int i = 0; i < num; i++)if (p&(1 << i))tim = (tim*(1 << cnt[i][x])) % mod;
			dp[p + x] = (dp[p + x] + tim*dp[p]) % mod;
		}
	}
	ll ans = 1;
	for (int i = 0; i < way; i++)ans = ans * 2 % mod;
	printf("%lld\n", (ans + mod - dp[(1 << num) - 1]) % mod);
}