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<bits/stdc++.h> 
using namespace std;
#define ll long long
#define C getchar()-48
inline ll read()
{
    ll s=0,r=1;
    char c=C;
    for(;c<0||c>9;c=C) if(c==-3) r=-1;
    for(;c>=0&&c<=9;c=C) s=(s<<3)+(s<<1)+c;
    return s*r;
}
#define find __builtin_popcount 
//这个函数是找一个数的2进制中有几个1 
const int N=15,M=1<<15,p=1e9+7; 
int n,m;
int g[N],dp[M],cf[M];
inline int add(int x,int y)//算两数相加模p意义下的值 
{
    return ((x+=y)>p)?x-p:x;
}
int main()
{
    n=read(),m=read();
    dp[0]=cf[0]=1;//初始状态 
    for(int i=1;i<=m;i++) cf[i]=(cf[i-1]<<1)%p;//预处理2的x次方 
    for(int i=1;i<=m;i++){int x=read()-1,y=read()-1;g[x]|=(1<<y);}//2进制存x能到达的边 
    for(int s=1,sed=1<<n;s<sed;s++) if((s&1)==(s>>1&1))//枚举s  因为1和2的点的状态相同所以直接判掉  下面同理 
    for(int u=s;u;u=u-1&s) if((u&1)==(u>>1&1))//枚举u 
    {
        int ans=1;
        for(int i=0;i<n;i++) if(s>>i&1)//枚举被选到点 
        {
            if(u>>i&1) ans=1ll*ans*cf[find(g[i]&(s^u))]%p;//假如为点集u的点 
            else ans=1ll*ans*(cf[find(g[i]&u)]-1)%p;//假如为点集t的点 
        }
        dp[s]=add(dp[s],1ll*dp[s^u]*ans%p);//转移 
    }
    cout<<(cf[m]-dp[(1<<n)-1]+p)%p;
    return 0;
}