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 fi first
#define se second
#define ll long long
#define dbg(v) cerr<<#v<<" = "<<v<<'\n'
#define vi vector<int>
#define vl vector <ll>
#define pii pair<int,int>
#define mp make_pair
#define db long double
#define pb push_back
#define all(s) s.begin(),s.end()
template < class T > T smin(T &a,T b) {if (a > b) a = b;return a;}
template < class T > T smax(T &a,T b) {if (a < b) a = b;return a;}
int D[1 << 17];
int P[1 << 17];
int s1[55],s2[55];
int W[1 << 17];
const int mod = 1e9 + 7;
int main(void)
{
    int n,m;
    cin>>n>>m;
    int p2 = 1;
    while (m --)
    {
        int a,b;
        cin>>a>>b;
        --a;--b;
        s1[a] |= 1 << b;
        s2[b] |= 1 << a;
        p2 = p2 * 2 % mod;
    }
    const int N = 1 << n;
    P[0] = 1;
    for (int mask = 1;mask < N;++mask)
        P[mask] = (P[mask - (mask & (-mask))] * 2) % mod;
    D[0] = 1;
    int ans = 0;
    for (int mask = 0;mask < N;++mask)
        if (D[mask])
        {
            const int rem = (N - 1) ^ mask;
            for (int sub = rem;sub;sub = (sub - 1) & rem)
            if ((sub&3) == 3 || !(sub&3))
            {
                const int R = rem ^ sub;
                int coef = D[mask];
                for (int j = 0;j < n;++j)
                    if ((R >> j) & 1)
                        coef = (1ll * coef * (P[s1[j] & sub] - 1)) % mod,coef = (1ll * coef * P[s2[j] & sub]) % mod;
                D[sub | mask] = (D[sub | mask] + coef) % mod;
            }
        }
    ans = (p2 + mod - D[N - 1]) % mod;
    cout << ans << '\n';
    return 0;
}
