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 maxn 16
#define maxs (1<<maxn)

typedef long long ll;
const ll mod = 1000000007;

ll dp[maxs];
int g[maxn];
int n, m;

int main()
{
    cin >> n >> m;
    for (int i = 1; i <= m; i++)
    {
        int x, y;
        cin >> x >> y;
        x--;
        y--;
        g[x] |= 1 << y;
    }
    dp[0] = 1;
    for (int s = 0; s < (1 << n); s++)
    {
        if (!dp[s])
        {
            continue;
        }
        int ns = ((1 << n) - 1)^s;
        for (int s1 = ns; s1; s1 = (s1 - 1)&ns)
        {
            if (s1 % 4 == 3)
            {
                continue;
            }
            ll x = dp[s];
            for (int k = 0; k < n; k++)
            {
                if (s & (1 << k))
                {
                    continue;
                }
                if (s1 & (1 << k))
                {
                    x = (x << __builtin_popcount(g[k] & (~(s | s1))));
                }
                else
                {
                    x = x * ((1 << __builtin_popcount(g[k] & s1)) - 1);
                }
                x %= mod;
            }
            dp[s | s1] += x;
            dp[s | s1] %= mod;
        }
    }
    printf("%lld\n", dp[(1 << n) - 1]);
    return 0;
}
