Problem: 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

Solution (in CPP):

Solution:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>

using namespace std;

const int MAXN = 15 + 2;
const int MAXM = 1 << 15;
const int MOD = 1000000000 + 7;

int fpow(int x, int p)
{
	int ret = 1;
	for(; p; p >>= 1, x = (long long)x * x % MOD)
		if(p & 1)
			ret = (long long)ret * x % MOD;
	return ret;
}

int n, m;
vector<int> edg[MAXN];
int cnt[MAXN][MAXM], f[MAXM];

void init()
{
	for(int i = 0; i < n; i++)
		for(int s = 0; s < (1 << n); s++)
		{
			cnt[i][s] = 1;
			for(int j = 0; j < edg[i].size(); j++)
				if((s >> edg[i][j]) & 1)
					cnt[i][s] = cnt[i][s] * 2 % MOD;
		}
}

int main()
{
	ios::sync_with_stdio(false);
	// freopen("1.in", "r", stdin);
	// freopen("1.out", "w", stdout);

	cin >> n >> m;
	for(int i = 1; i <= m; i++)
	{
		int x, y;
		cin >> x >> y;
		x -= 1, y -= 1;
		edg[x].push_back(y);
	}
	init();
	f[0] = 1;
	for(int i = 1; i < (1 << n); i++)
		if(i % 4 == 0 || i % 4 == 3)
		{
			f[i] = 0;
			for(int s = (i - 1) & i; s != i; s = (s - 1) & i)
			{
				int t = i ^ s;
				if((s & 1) ^ ((s >> 1) & 1))
					continue;
				// cerr << s << endl;
				int tmp = f[s];
				for(int j = 0; j < n; j++)
					if((s >> j) & 1)
						tmp = (long long)tmp * (cnt[j][t] - 1 + MOD) % MOD;
					else if((t >> j) & 1)
						tmp = (long long)tmp * cnt[j][s] % MOD;
				f[i] = (f[i] + tmp) % MOD;
			}
			// cerr << i << ' ' << f[i] << endl;
		}

	int ans = (fpow(2, m) - f[(1 << n) - 1] + MOD) % MOD;

	cout << ans << endl;


	return 0;
}