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<cstdio> 
#include<algorithm>
#include<cstring>
#include<cmath>
#include<set>
#include<map>
using namespace std;
#define fo(i,j,k) for(i=j;i<=k;i++)
#define fd(i,j,k) for(i=j;i>=k;i--)
#define cmax(a,b) (a=(a>b)?a:b)
#define cmin(a,b) (a=(a<b)?a:b)
typedef long long ll;
const int N=1e6+5,M=2e6+5,mo=1e9+7;
int n,m,x,y,ts,s1,s,ns,p,ans,f[1<<16],cnt[16][1<<16],i;
ll inc;
int tt,b[1000],nxt[1000],fst[1000];
void cr(int x,int y)
{
	tt++;
	b[tt]=y;
	nxt[tt]=fst[x];
	fst[x]=tt;
}
int main()
{
	scanf("%d %d",&n,&m);
	fo(i,1,m)
	{
		scanf("%d %d",&x,&y);
		cr(x,y);
	}
	ts=1<<n;ts--;
	fo(i,1,n) fo(s,0,ts) 
		for(p=fst[i];p;p=nxt[p]) if ((1<<b[p]-1)&s) cnt[i][s]++;
	fo(s,0,ts)
	if ((s&1)==((s>>1)&1))
	{
		f[s]=1;
		for(ns=(s-1)&s;ns;ns=(ns-1)&s)
			if ((ns&1)==((ns>>1)&1))
			{
				s1=s-ns;
				inc=f[s1];
				fo(i,1,n) 
				if (s1&(1<<i-1))
					(inc*=(1<<cnt[i][ns])-1)%=mo;
				else if (ns&(1<<i-1))
					(inc<<=cnt[i][s1])%=mo;
				f[s]=(f[s]+inc)%mo;
			}
	}
	ans=1;
	fo(i,1,m) ans=ans*2%mo;
	ans-=f[ts];
	ans=(ans+mo)%mo;
	printf("%d",ans);
}