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 = 310,MOD = 1e9+7;
int f[N][N][N],l[N],r[N],x[N],ans;
vector<int> p[N];
bool check(int i,int j,int k)
{
	vector<int>::iterator it;
	for(it = p[i].begin();it != p[i].end();++ it)
	{
		int cnt=1;
		if(j >= l[*it]) ++ cnt;
		if(k >= l[*it]) ++ cnt;
		if(cnt != x[*it]) return false;
	}
	return true;
}
int main()
{
	int n,m;
	scanf("%d %d",&n,&m);
	for(int i = 1;i <= m;++ i)
	{
		scanf("%d %d %d",&l[i],&r[i],&x[i]);
		p[r[i]].push_back(i);
	}
	if(check(1,0,0)) f[1][0][0] = 1;
	for(int i = 2;i <= n;++ i)
	{
		for(int j = 0;j <= i - 1;++ j)
		{
			for(int k = 0; k <= j;++ k)
			{
				if(j == k && j) continue;
				if(!check(i,j,k)) continue;
				f[i][j][k]=f[i - 1][j][k];
				if(j == i - 1||k == i - 1)
				{
					for(int l = 0 ;l <= i-1;++ l)
					{
						if(l < k) f[i][j][k] = (f[i][j][k] + f[j][k][l]) % MOD;
						else f[i][j][k] = (f[i][j][k] + f[j][l][k]) % MOD;
					}
				}
			}
		}
	}
	for(int i = 0;i < n;++ i)
		for(int j = 0;j < i;++ j)
			ans = (ans + f[n][i][j]) % MOD;
	ans = 6ll * ans % MOD;
	ans = (ans + f[n][0][0] * 3) % MOD;
	cout << ans;
}