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 <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;

const int mod=1000000007;

int pow2[110],inv2[110],f[20],s[1<<15],a[15],dp[1<<15][2];

int main()
{
	int n,m;scanf("%d%d",&n,&m);
	pow2[0]=1;for (int i=1;i<=max(n,m);i++) pow2[i]=pow2[i-1]*2%mod;
	inv2[0]=1;for (int i=1;i<=max(n,m);i++) inv2[i]=ll(inv2[i-1])*((mod+1)/2)%mod;
	for (int i=1;i<pow2[n];i++) s[i]=s[i>>1]+(i&1);
	for (int i=0;i<m;i++)
	{
		int x,y;scanf("%d%d",&x,&y);
		x--;y--;
		f[x]|=pow2[y];
	}
	dp[0][0]=1;
	for (int i=0;i<(1<<n)-1;i++)
	{
		int l=0;
		for (int j=0;j<n;j++)
			if (!(i&(1<<j))) a[l++]=j;
		for (int k=0;k<2;k++)
			if (dp[i][k])
				for (int j=1;j<pow2[l];j++)
				{
					int t=dp[i][k],x=0;bool ok=1;
					for (int p=0;p<l;p++) if (j&pow2[p]) x|=pow2[a[p]];
					for (int p=0;p<l;p++)
						if (j&pow2[p])
							t=ll(t)*inv2[s[f[a[p]]&x]]%mod;
						else
						{
							if (!(f[a[p]]&x)) {ok=0;break;}
							t=ll(t)*(pow2[s[f[a[p]]&x]]-1)%mod*inv2[s[f[a[p]]&x]]%mod;
						}
					if (ok)
					{
						bool tk=k|(l>=2&&a[0]==0&&a[1]==1&&((x&3)==3));
						dp[i|x][tk]=(dp[i|x][tk]+t)%mod;
					}
				}
	}
	printf("%d\n",(ll(mod+1-dp[pow2[n]-1][1])*pow2[m]%mod));
	return 0;
}
