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 <string>
#include <algorithm>
#define rep(i,st,ed) for (int i=st;i<=ed;++i)
#define lowbit(x) (x&-x)

typedef long long LL;
const int MOD=1e9+7;
const int N=18;

LL f[1<<N],g[N][1<<N],bin[155];

int read() {
	int x=0,v=1; char ch=getchar();
	for (;ch<'0'||ch>'9';v=(ch=='-')?(-1):(v),ch=getchar());
	for (;ch<='9'&&ch>='0';x=x*10+ch-'0',ch=getchar());
	return x*v;
}

int main(void) {
	bin[0]=1; rep(i,1,125) bin[i]=bin[i-1]*2LL%MOD;
	int n=read(),m=read();
	rep(i,1,m) {
		int x=read()-1,y=read()-1;
		++g[x][1<<y];
	}
	rep(i,0,n-1) {
		for (int j=1;j<(1<<n);++j) {
			g[i][j]=g[i][j-lowbit(j)]+g[i][lowbit(j)];
		}
	}
	for (int S=1;S<(1<<n);++S) {
		if ((S&1)!=((S>>1)&1)) continue;
		f[S]=1;
		for (int a=S,b,w;a;a=(a-1)&S) {
			if ((a&1)!=((a>>1)&1)) continue;
			b=S-a; w=f[b];
			rep(i,0,n-1) {
				if ((b>>i)&1) w=(bin[g[i][a]]-1)*w%MOD;
				if ((a>>i)&1) w=bin[g[i][b]]*w%MOD;
			}
			f[S]=(f[S]+w)%MOD;
		}
	}
	LL ans=(bin[m]+MOD-f[(1<<n)-1])%MOD;
	printf("%lld\n", ans);
	return 0;
}