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<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<bitset>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<climits>
#include<complex>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;

const int maxn = 17;
const int maxm = 300;
const int maxk = 270000;
const ll Mod = 1e9+7;

ll pw(ll x,int k)
{
	ll re=1ll;
	for(;k;k>>=1,x=x*x%Mod) if(k&1)
		re=re*x%Mod;
	return re;
}
ll pw2[maxm];
int n,m,d[maxn];
struct edge{int y,nex;}a[maxm]; int len,fir[maxn];
inline void ins(const int x,const int y){a[++len]=(edge){y,fir[x]};fir[x]=len;}

ll f[maxk],g[maxn][maxk],g2[maxn][maxk];

int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		int x,y; scanf("%d%d",&x,&y);
		ins(x,y); d[x]++;
	}
	pw2[0]=1ll; for(int i=1;i<maxm;i++) pw2[i]=pw2[i-1]*2ll%Mod;
	
	int al=1<<n;
	for(int i=1;i<=n;i++)
	{
		for(int j=0;j<al;j++)
		{
			int num=0;
			for(int k=fir[i],y=a[k].y;k;k=a[k].nex,y=a[k].y) if(j>>y-1&1) ++num;
			g[i][j]=pw2[num]-1ll;
		}
	}
	f[0]=1ll;
	for(int i=1;i<al;i++) if((i&3)==0||(i&3)==3)
	{
		for(int j=(i-1)&i;;j=(j-1)&i) if(f[j])
		{
			int oth=i-j; // sg = zero
			ll now=1ll;
			for(int k=1;k<=n;k++) if(j>>k-1&1)
				(now*=g[k][oth])%=Mod;
			for(int k=1;k<=n;k++) if(oth>>k-1&1)
				(now*=g[k][j]+1ll)%=Mod;
			(f[i]+=now*f[j]%Mod)%=Mod;
			if(!j) break;
		}
	}
	ll ans=pw2[m]-f[al-1];
	printf("%lld\n",(ans+Mod)%Mod);
	
	return 0;
}