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;
typedef long double ld;
typedef pair<int, int> pii;
const int oo = 1e9;
const int MOD = 1000000007;
const int BASE = 26;
const int N = 202010;
const int MX = 200000;
#define x first
#define y second
#define pb push_back
#define mp make_pair

int n , m, dp[310][310][310],l,r,x;

vector<pii> ad[310];

int calc(int r , int g , int b){
	int idx = max(r,max(g,b));
	if(idx == n+1)return 1;
	int &ret = dp[r][g][b];
	if(ret != -1)return ret;
	bool valid = true;
	for (int i = 0; i < ad[idx].size(); ++i){
		int v = ad[idx][i].y;
		int cnt = (r>=v)+(g>=v)+(b>=v);
		if(cnt != ad[idx][i].x){
			ret = 0;
			return ret;
		}
	}
	ret = (1ll*calc(idx+1,g,b)+calc(r,idx+1,b)+calc(r,g,idx+1))%MOD;
	return ret;
}
int fast(int a , int b){
	if(!b)return 1ll;
	int res = fast(a,b/2);
	res = (1ll*res*res)%MOD;
	if(b&1)return (1ll*res*a)%MOD;
	return res;
}
int main() {
	memset(dp,-1,sizeof dp);
	scanf("%d%d",&n,&m);
	for (int i = 0; i < m; ++i){
		scanf("%d%d%d",&l,&r,&x);
		ad[r].pb(mp(x,l));
	}
	ll res = (1ll*calc(0,0,0)*fast(3,MOD-2))%MOD;
	cout << res << endl;

	return 0;
}