Problem: There are N squares arranged in a row. The squares are numbered 1, 2, ..., N, from left to right.

Snuke is painting each square in red, green or blue. According to his aesthetic sense, the following M conditions must all be satisfied. The i-th condition is:

* There are exactly x_i different colors among squares l_i, l_i + 1, ..., r_i.



In how many ways can the squares be painted to satisfy all the conditions? Find the count modulo 10^9+7.

Constraints

* 1 ≤ N ≤ 300
* 1 ≤ M ≤ 300
* 1 ≤ l_i ≤ r_i ≤ N
* 1 ≤ x_i ≤ 3

Input

Input is given from Standard Input in the following format:


N M
l_1 r_1 x_1
l_2 r_2 x_2
:
l_M r_M x_M


Output

Print the number of ways to paint the squares to satisfy all the conditions, modulo 10^9+7.

Examples

Input

3 1
1 3 3


Output

6


Input

4 2
1 3 1
2 4 2


Output

6


Input

1 3
1 1 1
1 1 2
1 1 3


Output

0


Input

8 10
2 6 2
5 5 1
3 5 2
4 7 3
4 4 1
2 3 1
7 7 1
1 5 2
1 7 3
3 4 2


Output

108

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

const int MOD = 1e9 + 7;

int dp[310][310][310];
vector<int> cond[310];

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n, m;
	cin >> n >> m;
	vector<int> l(m), r(m), x(m);
	for (int i = 0; i < m; i++) {
		cin >> l[i] >> r[i] >> x[i];
		cond[r[i]].push_back(i);
	}
	dp[0][0][0] = 1;
	for (int i = 0; i <= n; i++) {
		for (int j = 0; j <= n; j++) {
			if (j != 0 && j == i) continue;
			for (int k = 0; k <= n; k++) {
				if (k != 0 && (k == i || k == j)) continue;
				int cmax = max({i, j, k});
				for (int idx : cond[cmax]) {
					int cnt = 0;
					cnt += (l[idx] <= i);
					cnt += (l[idx] <= j);
					cnt += (l[idx] <= k);
					if (cnt != x[idx]) dp[i][j][k] = 0;
				}
				(dp[cmax + 1][j][k] += dp[i][j][k]) %= MOD;
				(dp[i][cmax + 1][k] += dp[i][j][k]) %= MOD;
				(dp[i][j][cmax + 1] += dp[i][j][k]) %= MOD;
			}
		}
	}

	int ans = 0;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			if (i != 0 && i == j) continue;
			(ans += dp[n][i][j]) %= MOD;
			(ans += dp[i][n][j]) %= MOD;
			(ans += dp[i][j][n]) %= MOD;
		}
	}
	cout << ans << endl;
	return 0;
}