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<stdio.h>
#include<algorithm>
using namespace std;
int n;
int dp[333][333][333];
int pos1L[333],pos2R[333],pos2L[333],pos3R[333];
#define mod 1000000007
void addto(int&x,int y){
	x+=y;
	if(x>=mod)x-=mod;
}
bool chk(int i,int y,int z){
	if(pos1L[i] <= y)return false;
	if(pos2R[i] != -1){
		if(!(pos2R[i] <= y))return false;
		if(!(pos2L[i] > z))return false;
	}
	if(pos3R[i] != -1){
		if(!(pos3R[i] <= z))return false;
	}
	return true;
}
int main(){
	int m;
	scanf("%d%d",&n,&m);
	for(int i=1; i<=n; i++){
		pos1L[i] = i;
		pos2R[i] = -1;
		pos2L[i] = i-1;
		pos3R[i] = -1;
	}
	while(m--){
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		if(z == 1){
			pos1L[y] = min(pos1L[y], x);
		}else
		if(z == 2){
			pos2R[y] = max(pos2R[y], x);
			pos2L[y] = min(pos2L[y], x);
		}else
			pos3R[y] = max(pos3R[y], x);
	}
	dp[1][0][0] = 3;
	int res=0;
	for(int i=1; i<=n; i++)
	for(int j=i-1; j>=0; j--)
	for(int k=max(j-1,0); k>=0; k--){
		if(!chk(i,j,k))
			dp[i][j][k]=0;
		if(i==n)addto(res, dp[i][j][k]);else{
			addto(dp[i+1][j][k], dp[i][j][k]);
			addto(dp[i+1][i][k], dp[i][j][k]);
			addto(dp[i+1][i][j], dp[i][j][k]);
		}
	}
	printf("%d\n",res);
	return 0;
}