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 Mod = 1e9 + 7;

int x, y, mask, m0, m1, res, dp[1 << 15], n, m, i, cntb[1 << 15], go[20], ans = 1;
bool good[1 << 15];

void inm(int &x, int y)
{
    x = (long long) x * y % Mod;
}

void add(int &x, int y)
{
    x += y;
    if(x >= Mod) x -= Mod;
}

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

    cin >> n >> m;
    for(i=1; i<=m; ++i)
    {
        cin >> x >> y;
        --x; --y;
        go[x] |= (1<<y);
        inm(ans, 2);
    }

    for(i = 0; i < (1<<n); ++i)
    {
        good[i] = ((i&3) == 3 || (i&3) == 0);
        cntb[i] = cntb[i>>1] + (i&1);
    }

    dp[0] = 1;
    for(mask = 1; mask < (1<<n); ++ mask)
    if(good[mask])
    {
        dp[mask] = 1;
        for(m1 = mask & (mask - 1); m1; m1 = mask & (m1 - 1))
        {
            m0 = mask ^ m1;
            if(!good[m0] || !good[m1]) continue;

            res = dp[m1];
            for(i=0; i<n; ++i)
                if(m0 & (1<<i))
                    inm(res, (1 << cntb[go[i] & m1]));
                else if(m1 & (1<<i))
                    inm(res, (1 << cntb[go[i] & m0]) - 1);
            add(dp[mask], res);
        }
    }

    add(ans, Mod - dp[mask-1]);
    cout << ans << '\n';

    return 0;
}
