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;
typedef long long ll;

template <typename _tp> inline void read(_tp&x){
	char ch=getchar(),ob=0;x=0;
	while(ch!='-'&&!isdigit(ch))ch=getchar();if(ch=='-')ob=1,ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();if(ob)x=-x;
}

const int N = 307, p = 1e9+7;
int f[N][N][N];
int n, m;

inline void pls(int&x, const int y) {x = x+y < p ? x+y : x+y-p;}

struct node {int l, x;};
vector <node> vec[N];
vector <node> :: iterator itr;

inline bool ill(int i, int j, int k) {
	int r = max(max(i, j), k);
	for(itr = vec[r].begin(); itr != vec[r].end(); ++ itr)
		if(itr->x != (i >= itr->l) + (j >= itr->l) + (k >= itr->l)) return true;
	return false;
}

int main() {
	read(n), read(m);
	for(int i=1,l,r,x;i<=m;++i) {
		read(l), read(r), read(x);
		vec[r].push_back((node) {l, x});
	}
	int Ans = 0;
	f[0][0][0] = 1;
	for(int i=0;i<=n;++i)
	for(int j=0;j<=n;++j)
	for(int k=0;k<=n;++k) if(f[i][j][k]) {
		if(ill(i, j, k)) {f[i][j][k] = 0; continue;}
		const int r = max(max(i, j), k), v = f[i][j][k];
		if(i != n) pls(f[r+1][j][k], v);
		if(j != n) pls(f[i][r+1][k], v);
		if(k != n) pls(f[i][j][r+1], v);
		if(r == n) pls(Ans, v);
	}
	printf("%d\n", Ans);
	return 0;
}