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:
//              +-- -- --++-- +-In the name of ALLAH-+ --++-- -- --+              \\

#include <bits/stdc++.h>

#define F first
#define S second
#define _sz(x) (int)x.size()

using namespace std ;
using ll = long long ;
using ld = long double ;
using pii = pair <int , int> ;

const int N = 300 + 10 , mod = 1e9 + 7 ;
int n , m , ans ;
int dp[N][N][N] ;
vector <pii> vec[N] ;

inline void add (int &a , int b) {
	a += b ;
	if (a >= mod) a -= mod ;
}

int main(){
	ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0) ;

	cin >> n >> m ;

	for (int i = 0 , l , r , x ; i < m ; i ++) {
		cin >> l >> r >> x ;
		vec[r].push_back({l , x}) ;
	}

	dp[0][0][0] = 1 ;
	for (int a = 0 ; a <= n ; a ++)
		for (int b = 0 ; b <= n ; b ++)
			for (int c = 0 ; c <= n ; c ++) {
				int i = max(max(a , b) , c) ;

				for (pii p : vec[i]) {
					int l = p.F , x = p.S ;

					int num = (a >= l) + (b >= l) + (c >= l) ;

					if (num != x) {
						dp[a][b][c] = 0 ;
						break ;
					}
				}

				add(dp[i + 1][b][c] , dp[a][b][c]) ;
				add(dp[a][i + 1][c] , dp[a][b][c]) ;
				add(dp[a][b][i + 1] , dp[a][b][c]) ;


				if (i == n) add(ans , dp[a][b][c]) ; 
		}

	cout << ans << '\n' ;
}