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>
#define f first
#define s second
using namespace std;
using ll = long long;
using ii = pair<int, int>;
constexpr ll MOD = 1e9+7;

inline void md(ll & a, ll & b) { a = (a + b) % MOD; }

bitset<303> valid[303][303];
ll dp[303][303][303];
vector<ii> C[303];

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0);

	int N, M; cin >> N >> M;
	for (int i = 0, l, r, x; i < M; ++i) {
		cin >> l >> r >> x; C[r].emplace_back(l, x);		
	}

	for (int r = 0; r <= N; ++r) for (int g = 0; g <= N; ++g) for (int b = 0; b <= N; ++b) {
		valid[r][g][b] = 1;
		int m = max({ r, g, b });
		for (auto & p : C[m]) {
			if ((p.f <= r || m == r) + (p.f <= g || m == g) + (p.f <= b || m == b) != p.s) valid[r][g][b] = 0;
		}
	}

	dp[0][0][0] = 1;
	for (int r = 0; r <= N; ++r) for (int g = 0; g <= N; ++g) for (int b = 0; b <= N; ++b) {
		if (!valid[r][g][b]) continue;
		int m = max({ r, g, b });
		md(dp[m + 1][g][b], dp[r][g][b]), md(dp[r][m + 1][b], dp[r][g][b]), md(dp[r][g][m + 1], dp[r][g][b]);
	}

	ll ans = 0;
	for (int r = 0; r <= N; ++r) for (int g = 0; g <= N; ++g) for (int b = 0; b <= N; ++b)
		if ((r == N || g == N || b == N) && valid[r][g][b]) md(ans, dp[r][g][b]);
	cout << ans;
}