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<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define mp(x,y) make_pair(x,y)
#define mod 1000000007
#define fi first
#define se second
#define MN 303
using namespace std;
inline int in(){
    int x=0;bool f=0;char c;
    for (;(c=getchar())<'0'||c>'9';f=c=='-');
    for (x=c-'0';(c=getchar())>='0'&&c<='9';x=(x<<3)+(x<<1)+c-'0');
    return f?-x:x;
}
struct st{
	int l,r,x;
}a[MN];
typedef pair<int,int> pr;
vector <pr> v[MN];
int f[MN][MN][MN];
int n,m,res;
inline bool cmp(st a,st b){
	return (a.r==b.r)?a.l<b.l:a.r<b.r;
}
int main()
{
	n=in();m=in();
	for (int i=1;i<=m;++i){
		a[i].l=in();a[i].r=in();a[i].x=in();
		v[a[i].r].push_back(mp(a[i].l,a[i].x));
	}f[1][0][0]=3;
	for (int i=1;i<=n;++i)
	for (int j=0;j<i||(!j);++j)
	for (int k=0;k<i||(!k);++k){
		int siz=v[i].size(),cnt=0;
		for (int h=0;h<siz;++h){
			cnt=(v[i][h].fi<=i)+(v[i][h].fi<=j)+(v[i][h].fi<=k);
			if (cnt!=v[i][h].se) f[i][j][k]=0;
		}
		f[i+1][j][k]=(1ll*f[i+1][j][k]+f[i][j][k])%mod;
		f[i+1][i][k]=(1ll*f[i+1][i][k]+f[i][j][k])%mod;
		f[i+1][i][j]=(1ll*f[i+1][i][j]+f[i][j][k])%mod;
	}
	for (int j=0;j<n;++j)
	for (int k=0;k<j||(!k);++k) res=(1ll*res+f[n][j][k])%mod;
	printf("%d",res);return 0;
}