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 <vector>
#include <algorithm>
using namespace std;
typedef pair<int,int>pii;
 
const int pps=1e9+7;
 
int n,m,ans;
vector<pii>q[305];
int f[305][305][305];
vector<pii>::iterator it;
 
int read() {
    int x=0,f=1;char ch=getchar();
    for(;ch<'0'||ch>'9';ch=getchar())if(ch=='-')f=-1;
    for(;ch>='0'&&ch<='9';ch=getchar())x=x*10+ch-'0';
    return x*f;
}
 
int main() {
    n=read(),m=read();
    for(int i=1;i<=m;i++) {
        int l=read(),r=read(),x=read();
        q[r].push_back(make_pair(l,x));
    }f[0][0][0]=1;
    for(int i=1;i<=n;i++) {
        for(int j=0;j<i;j++)
            for(int k=0;k<=j;k++) {//<=是因为j,k可以全部等于0
                f[i][j][k]=(f[i][j][k]+f[i-1][j][k])%pps;
                f[i][i-1][k]=(f[i][i-1][k]+f[i-1][j][k])%pps;
                f[i][i-1][j]=(f[i][i-1][j]+f[i-1][j][k])%pps;
                for(it=q[i].begin();it!=q[i].end();it++) {
                    int tmp=(j>=(*it).first)+(k>=(*it).first)+1;
                    if(tmp!=(*it).second)f[i][j][k]=0;
                }
            }
    }
    for(int i=0;i<n;i++)
        for(int j=0;j<=i;j++)
            ans=(ans+f[n][i][j])%pps;
    printf("%d\n",ans);
    return 0;
}