p03717 AtCoder Regular Contest 074 - RGB Sequence
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
```python
N,M=map(int,input().split())
q=[[] for i in range(N+1)]
for i in range(M):
    l,r,x=map(int,input().split())
    q[r].append((l,x))
mod=10**9+7

dp=[[0 for i in range(N+1)] for i in range(N+1)]

i=N
for j in range(N+1):
    for k in range(N+1):
        check=7
        for l,x in q[i]:
            if i==l:
                check=(x==1)*check
            elif i-1>=l>j:
                check=(x==1)*(check%2)+(x==2)*(check&6)
            elif j>=l>k:
                check=(x==2)*(check&3)+(x==3)*(check&4)
            else:
                check*=(x==3)
        dp[j][k]=bin(check).count("1")

ndp=[[0 for j in range(N+1)] for k in range(N+1)]
for i in range(N-1,0,-1):
    for j in range(i):
        for k in range(j+1):
            check=7
            for l,x in q[i]:
                if i==l:
                    check=(x==1)*check
                elif i-1>=l>j:
                    check=(x==1)*(check%2)+(x==2)*(check&6)
                elif j>=l>k:
                    check=(x==2)*(check&3)+(x==3)*(check&4)
                else:
                    check*=(x==3)
            ndp[j][k]=(dp[j][k]*((check//1)%2)+dp[i-1][k]*((check//2)%2)+dp[i-1][j]*((check//4)%2))%mod

    dp=ndp

print(dp[0][0])
```