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;

int N, M, msk[15], dp[1 << 15], nrb[1 << 15];
bool ok[1 << 15];
const int mod = 1e9 + 7;
const int long long mod2 = 1LL * mod * mod;
void mul (int &x, int y) {x = (1LL * x * y) % mod;}
void adto (int &x, int y) {x += y; if (x >= mod) x -= mod;}

int main ()
{
//freopen ("input", "r", stdin);
//freopen ("output", "w", stdout);

scanf ("%d %d", &N, &M);
int ans = 1;
while (M --)
{
    int x, y;
    scanf ("%d %d", &x, &y), x --, y --;
    msk[x] |= 1 << y, ans += ans;
    if (ans >= mod) ans -= mod;
}
for (int i=0; i<1 << N; i++)
{
    nrb[i] = nrb[i >> 1] + (i & 1);
    ok[i] = (i & 1) ^ ((i >> 1) & 1) ^ 1;
}
for (int i=0; i<1 << N; i++)
if (ok[i])
{
    dp[i] = 1;
    for (int j=(i - 1) & i; j != 0; j = (j - 1) & i)
    if (ok[j] && ok[j ^ i])
    {
        int cnt = dp[j], j0 = j ^ i;
        for (int k=0; k<N; k++)
            if ((j >> k) & 1)
                mul (cnt, (1 << nrb[msk[k] & j0]) - 1);
        for (int k=0; k<N; k++)
            if ((j0 >> k) & 1)
                mul (cnt, 1 << nrb[msk[k] & j]);
        dp[i] += cnt;
        if (dp[i] >= mod)
            dp[i] -= mod;
    }
}
ans -= dp[(1 << N) - 1];
if (ans < 0) ans += mod;
printf ("%d\n", ans);

return 0;
}
