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 PYTHON3):

Solution:
import sys
input = sys.stdin.readline

mod = 10**9+7



# (i, j, k+1)に置くのが、条件(l,x)に適しているか
def check(i, j, l, x):
    return (x == 3 and l <= i) or (x == 2 and i < l <= j) or (x == 1 and j < l)


N, M = map(int, input().split())
LRX = [list(map(int, input().split())) for _ in range(M)]

LX = [[] for _ in range(N+2)]
for l, r, x in LRX:
    LX[r].append((l, x))

dp = [[[0]*(j+1) for j in range(i+1)] for i in range(N+2)]
# dp[i][j][k] (k <= j <= i) 
# 三色それぞれ、最後に使ったindex(1-indexed)がi,j,k

dp[0][0][0] = 1

for i in range(N+1):
    for j in range(i, N+1):
        for k in range(j, N+1):
            
            # 一番前の色を置く
            ok = True
            for l, x in LX[k+1]:
                if not check(j, k, l, x):
                    ok = False
                    break
            if ok:
                dp[k+1][k][j] = (dp[k+1][k][j] + dp[k][j][i]) % mod
            
            # 最後
            ok = True
            for l, x in LX[k+1]:
                if not check(i, j, l, x):
                    ok = False
                    break
            if ok:
                dp[k+1][j][i] = (dp[k+1][j][i] + dp[k][j][i]) % mod
            
            # 二番目
            ok = True
            for l, x in LX[k+1]:
                if not check(i, k, l, x):
                    ok = False
                    break
            if ok:
                dp[k+1][k][i] = (dp[k+1][k][i] + dp[k][j][i]) % mod
        

ans = 0
for a in range(N+1):
    for b in range(a,N+1):
        ans = (ans + dp[N][b][a]) % mod

print(ans)