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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;
const int MAXN = 305;
const int MOD = 1000000007;
vector < pair<int,int> > A[MAXN];
int dp[MAXN][MAXN][MAXN];
void mod_add(int &x, int y)
{
	x = (x + y)%MOD;
}
int main()
{
	int n,m;
	scanf("%d %d", &n, &m);
	for (int i = 0; i < m; ++i)
	{
		int ty, l, r;
		scanf("%d %d %d", &l, &r, &ty);
		A[r].push_back(make_pair(l,ty));
	}
	dp[0][0][0] = 1;
	int ans = 0;
	for (int i = 0; i <= n; ++i)
	{
		for (int j = 0; j <= n; ++j)
		{
			for (int k = 0; k <= n; ++k)
			{
				int pos[] = {i,j,k};
				sort(pos, pos+3);
				for (int ii = 0; ii < A[pos[2]].size(); ++ii)
				{
					int l = A[pos[2]][ii].first;
					int num = A[pos[2]][ii].second;
					if(num == 1 && pos[1] >= l)
						dp[i][j][k] = 0;
					if(num == 2 && (pos[0] >= l || pos[1] < l))
						dp[i][j][k] = 0;
					if(num == 3 && pos[0] < l)
						dp[i][j][k] = 0;
				}
				mod_add(dp[pos[2] + 1][j][k], dp[i][j][k]);
				mod_add(dp[i][pos[2] + 1][k], dp[i][j][k]);
				mod_add(dp[i][j][pos[2] + 1], dp[i][j][k]);
				if(pos[2] == n)
					mod_add(ans, dp[i][j][k]);
			}
		}
	}
	printf("%d\n", ans);
	return 0;
}