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;
const int N = 20;
const int M = 100000;
const int MOD = 1e9 + 7;
int n, m;
int A[N], B[N], S = 1;
int f[M], c[M];
void ADD(int &t, int d)
{
    t += d; if (t >= MOD) t -= MOD;
}
int main()
{
    cin >> n >> m;
    for (int i = 1; i <= m; ++ i)
    {
        int a, b;
        cin >> a >> b;
        B[a - 1] |= 1 << (b - 1);
    }
    for (int i = 1; i < (1 << n); ++ i)
        c[i] = c[i >> 1] + (i & 1);
    f[(1 << n) - 1] = 1;
    for (int i = (1 << n) - 1; i; -- i)
    {
        for (int j = i; j; j = (j - 1) & i)
        {
            if (!(((bool)(j & 1)) ^ ((bool)(j & 2))))
            {
                int w = f[i];
                for (int k = 0; k < n; ++ k)
                    if ((1 << k) & (i ^ j))
                        w = 1ll * w * ((1 << c[B[k] & j]) - 1) % MOD;
                for (int k = 0; k < n; ++ k)
                    if ((1 << k) & j)
                        w = 1ll * w * (1 << c[B[k] & (i ^ j)]) % MOD;
                ADD(f[i ^ j], w);
            }
        }
    }
    for (int i = 0; i < m; ++ i) S = 2ll * S % MOD;
    cout << (S - f[0] + MOD) % MOD << endl;
}
