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<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define rep(i,a,b) for(i=a;i<=b;i++)
using namespace std;
const int mod=1e9+7;
const int N=(1<<16)+5;
int dp[N],to[16],from[16],Log[N],ok[N];int pow[300];int num[N];

inline int lowbit(int x){return x&(-x);}

int main(){
	//freopen("a.in","r",stdin);
	//freopen("a.out","w",stdout);
	register int i,S,x,y,cnt;
	int totS;int n,m;
	scanf("%d%d",&n,&m);totS=(1<<n)-1;
	rep(i,0,n)	Log[1<<i]=i;
	rep(i,1,totS)	ok[i]=((i&1)+(i>>1&1))&1,ok[i]^=1,dp[i]=ok[i];
	pow[0]=1;rep(i,1,m)	pow[i]=1ll*pow[i-1]*2%mod;
	rep(i,1,totS)	num[i]=num[i>>1]+(i&1);
	rep(i,1,m){
		scanf("%d%d",&x,&y);x--;y--;
		to[x]|=(1<<y);from[y]|=1<<x;
	}
	//rep(i,1,totS)	cerr<<ok[i]<<' ';cerr<<endl;
	rep(i,1,totS) if(ok[i]){
		dp[i]=1;
		for(S=(i-1)&i;S;S=(S-1)&i) if(ok[S]){
			x=i^S;cnt=dp[x];
			while(x){
				y=lowbit(x);
				cnt=1ll*cnt*(pow[num[to[Log[y]]&S]]-1)%mod;cnt=1ll*cnt*pow[num[from[Log[y]]&S]]%mod;
				x-=y;
			}
			dp[i]=(dp[i]+cnt)%mod;
		}
	}
	//rep(i,1,totS)	cerr<<dp[i]<<' ';cerr<<endl;
	//rep(i,1,totS)	cerr<<f[i]<<' ';cerr<<endl;
	printf("%d\n",(pow[m]-dp[totS]+mod)%mod);
}