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 N = 305;
const int md = 1e9 + 7;

int n, m;
vector < pair<int,int> > seg[N];
int dp[N][N][N];

void add(int &x, int y) { x += y; while(x >= md) x -= md; if (x < 0) x += md; }

bool check(int i, int j, int k) {
	for (auto segment : seg[i]) {
		int lef = segment.first; // left point
		int dif = segment.second; // different colors
		int cnt = (i >= lef) + (j >= lef) + (k >= lef);
		if (cnt != dif) return false;
	}
	return true;
}

int main() {
	ios_base::sync_with_stdio(false); cin.tie(0);
	cin >> n >> m;
	for (int i = 1; i <= m; ++i) {
		int l, r, x; cin >> l >> r >> x;
		seg[r].push_back(make_pair(l, x));
	}
	dp[0][0][0] = 1;
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j < i; ++j) {
			for (int k = 0; k <= j; ++k) {
				if (!dp[i-1][j][k]) continue;
				int nj, nk;
				// i-1
				if (check(i, j, k)) add(dp[i][j][k], dp[i-1][j][k]);
				// j
				if (check(i, i-1, k)) add(dp[i][i-1][k], dp[i-1][j][k]);
				// k
				if (check(i, i-1, j)) add(dp[i][i-1][j], dp[i-1][j][k]);
			}
		}
	}	

	int ans = 0;
	for (int j = 0; j < n; ++j) {
		for (int k = 0; k <= j; ++k) {
			add(ans, dp[n][j][k]);
		}
	}
	printf("%lld\n", ans);
}