Problem:
Score : 800 points

Problem Statement

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.

Sample Input 1

3 1
1 3 3

Sample Output 1

6

The six ways are:

RGB

RBG

GRB

GBR

BRG

BGR

where R, G and B correspond to red, green and blue squares, respectively.

Sample Input 2

4 2
1 3 1
2 4 2

Sample Output 2

6

The six ways are:

RRRG

RRRB

GGGR

GGGB

BBBR

BBBG

Sample Input 3

1 3
1 1 1
1 1 2
1 1 3

Sample Output 3

0

There are zero ways.

Sample Input 4

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

Sample Output 4

108

Solution in C++ (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;

typedef long long LL;

inline int read()
{
	char c=getchar();int num=0,f=1;
	for(;!isdigit(c);c=getchar())
		f=c=='-'?-1:f;
	for(;isdigit(c);c=getchar())
		num=num*10+c-'0';
	return num*f;
}

const int N=305;
const int mod=1e9+7;

inline void add(int &x,int y){(x+=y)>=mod?x-=mod:0;}

int n,m;
int f[N][N][N];
vector<pii> vec[N];

bool check(int r,int g,int b)
{
	int k=max(r,max(g,b));
	int siz=vec[k].size();
	for(int i=0,l,x,cnt;i<siz;++i)
	{
		l=vec[k][i].fi,x=vec[k][i].se;
		cnt=0;
		if(r>=l) ++cnt;
		if(g>=l) ++cnt;
		if(b>=l) ++cnt;
		if(cnt!=x) return 0;
	}
	return 1;
}

int main()
{
	n=read(),m=read();
	for(int i=1,l,r,x;i<=m;++i)
	{
		l=read(),r=read(),x=read();
		vec[r].pb(mp(l,x));
	}
	f[0][0][0]=1;
	int ans=0;
	for(int r=0;r<=n;++r)
		for(int g=0;g<=n;++g)
			for(int b=0;b<=n;++b)
			{
				if(!f[r][g][b]) continue;
				if(!check(r,g,b)){f[r][g][b]=0;continue;}
				int k=max(r,max(g,b));
				if(k==n) add(ans,f[r][g][b]);
				add(f[k+1][g][b],f[r][g][b]),
				add(f[r][k+1][b],f[r][g][b]),
				add(f[r][g][k+1],f[r][g][b]);
			}
	cout<<ans;
	return 0;
}