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<bits/stdc++.h>
using namespace std;

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

const int mod=1000000007;
inline void add(int &a,int b){
    a+=b;
    if(a>=mod)a-=mod;
}


int pc[1<<15];

int N,M;
int G[15];
int rG[15];
int dp[1<<15];

signed main(){
    cin>>N>>M;
    rep(i,M){
        int x,y;
        cin>>x>>y;
        x--;y--;
        G[x]|=1<<y;
        rG[y]|=1<<x;
    }

    rep(i,1<<15)pc[i]=__builtin_popcount(i);

    dp[0]=1;
    rep(i,1<<N){
        int sup=(1<<N)-1-i;
        int sub=sup;
        do{
            int O=(1<<N)-1-i-sub;
            if(sub!=0&&(sub&3)!=3){
                int w=1;
                rep(j,N){
                    if(i>>j&1){
                        w=w*(1<<pc[G[j]&sub])%mod;
                    }
                    else if(O>>j&1){
                        w=w*((1<<pc[G[j]&sub])-1)%mod;
                    }
                }

                add(dp[i|sub],dp[i]*w%mod);
            }
            sub=(sub-1)&sup;
        }while(sub!=sup);
    }


    cout<<dp[(1<<N)-1]<<endl;
    return 0;
}