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<vector>
using namespace std;

typedef long long ll;
const int MAXN=20;
const int MOD=1e9+7;

int n,m;
int cnt[MAXN][1<<MAXN],d[1<<MAXN];
vector<int> E[MAXN];

int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++){
        int u,v;
        scanf("%d%d",&u,&v);
        E[u].push_back(v);
    }
    int tot=(1<<n)-1;
    for(int i=1;i<=n;i++)
        for(int j=0;j<=tot;j++)
            for(int k=0;k<(int)E[i].size();k++){
                int v=E[i][k];
                if(j&(1<<(v-1))) cnt[i][j]++;
            }
    for(int s=0;s<=tot;s++)
        if((s&1)==((s>>1)&1)){
            d[s]=1;
            for(int ns=(s-1)&s;ns;ns=s&(ns-1))
                if((ns&1)==((ns>>1)&1)){
                    int s1=s-ns;
                    ll inc=d[s1];
                    for(int i=1;i<=n;i++)
                        if(s1&(1<<(i-1)))
                            (inc*=(1<<cnt[i][ns])-1)%=MOD;
                        else if(ns&(1<<(i-1)))
                            (inc<<=cnt[i][s1])%=MOD;
                    d[s]=(d[s]+inc)%MOD;
                }
        }
    ll ans=1;
    for(int i=1;i<=m;i++) ans=ans*2%MOD;
    ans-=d[tot];
    ans=(ans+MOD)%MOD;
    printf("%lld",ans);
}
