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<bits/stdc++.h>
#define ll long long
using namespace std;
const int mod=1e9+7;
ll f[1<<17];
int G[20],num[1<<17];
int read()
{
	int x=0;char c=getchar(),flag='+';
	while(!isdigit(c)) flag=c,c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return flag=='-'?-x:x;
}
int main()
{
	int n=read(),m=read();
	for(int i=1;i<=(1<<n);i++) num[i]=num[i>>1]+(i&1);
	//for(int i=1;i<=10;i++) cout<<num[i]<<' ';
	for(int i=1;i<=m;i++)
	{
		int u=read(),v=read();
		G[u]|=1<<v-1;
	}
	f[0]=1;
	for(int s=1;s<(1<<n);s++)
	{
		if((s&1)==((s>>1)&1))
		{
			for(int u=s;u;u=(u-1)&s)
			{
				if((u&1)==((u>>1)&1))
				{
					ll tmp=1;
					int t=s^u;
					for(int j=1;j<=n;j++)
					{
						if(s&(1<<j-1))
						{
							if(u&(1<<j-1)) tmp=tmp*(1<<num[G[j]&t])%mod;
							else tmp=tmp*((1<<(num[G[j]&u]))-1)%mod;
						}
					}
					f[s]=(f[s]+tmp*f[t])%mod;
				}
			}
		}
	}
	ll ans=1;
	for(int i=1;i<=m;i++) ans=ans*2%mod;
	cout<<(ans-f[(1<<n)-1]+mod)%mod;
	return 0;
}
