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>
using namespace std;
#define int long long
#define N 20
#define M 1000000007
int he[N],ne[N*N],e[N*N],cnt,n,m;
int f[1<<N],lg[1<<N],pw[N*N],ans;
void add(int u,int v){ne[++cnt]=he[u];he[u]=cnt;e[cnt]=v;}
signed main(){
    scanf("%lld%lld",&n,&m);
    for (int i=1;i<=m;i++){
        int u,v; scanf("%lld%lld",&u,&v); u--; v--;
        add(u,v);
    }
    pw[0]=1;
    for (int i=1;i<=m;i++) pw[i]=(pw[i-1]<<1)%M;
    lg[1]=0;
    for (int i=2;i<1<<n;i++) lg[i]=lg[i>>1]+1;
    f[0]=1;
    for (int i=1;i<1<<n;i++){
        f[i]=1; 
        int t=(i&1)+((i&2)>>1);
        if (t==1){f[i]=0; continue;}
        for (int j=(i-1)&i;j;j=(j-1)&i){
        	int t=(j&1)+((j&2)>>1);
        	if (t==1) continue;
            int p=i^j; bool flag=1; int sum=1;
            for (;p;p-=p&-p){
                int x=p&-p;
                x=lg[x]; int tot=0;
                for (int k=he[x];k;k=ne[k]){
                    if ((1<<e[k]) &j){
                        tot++;
                    }
                }
                if (!tot) {flag=0; break;}
                (sum*=(pw[tot]-1))%=M; 
            }
            if (!flag) continue;
            p=j;
            for (;p;p-=p&-p){
                int x=p&-p;
                x=lg[x]; int tot=0;
                for (int k=he[x];k;k=ne[k])
                    if ((1<<e[k]) &(i^j)) tot++;
                (sum*=pw[tot])%=M;
            }
            (f[i]+=f[i^j]*sum%M)%=M;
        }
    }
    printf("%lld\n",(pw[m]-f[(1<<n)-1]+M)%M);
}