Problem: There are N people standing on the x-axis. Let the coordinate of Person i be x_i. For every i, x_i is an integer between 0 and 10^9 (inclusive). It is possible that more than one person is standing at the same coordinate.

You will given M pieces of information regarding the positions of these people. The i-th piece of information has the form (L_i, R_i, D_i). This means that Person R_i is to the right of Person L_i by D_i units of distance, that is, x_{R_i} - x_{L_i} = D_i holds.

It turns out that some of these M pieces of information may be incorrect. Determine if there exists a set of values (x_1, x_2, ..., x_N) that is consistent with the given pieces of information.

Constraints

* 1 \leq N \leq 100 000
* 0 \leq M \leq 200 000
* 1 \leq L_i, R_i \leq N (1 \leq i \leq M)
* 0 \leq D_i \leq 10 000 (1 \leq i \leq M)
* L_i \neq R_i (1 \leq i \leq M)
* If i \neq j, then (L_i, R_i) \neq (L_j, R_j) and (L_i, R_i) \neq (R_j, L_j).
* D_i are integers.

Input

Input is given from Standard Input in the following format:


N M
L_1 R_1 D_1
L_2 R_2 D_2
:
L_M R_M D_M


Output

If there exists a set of values (x_1, x_2, ..., x_N) that is consistent with all given pieces of information, print `Yes`; if it does not exist, print `No`.

Examples

Input

3 3
1 2 1
2 3 1
1 3 2


Output

Yes


Input

3 3
1 2 1
2 3 1
1 3 5


Output

No


Input

4 3
2 1 1
2 3 5
3 4 2


Output

Yes


Input

10 3
8 7 100
7 9 100
9 8 100


Output

No


Input

100 0


Output

Yes

Solution (in PYTHON3):

Solution:
import sys
from collections import deque
sys.setrecursionlimit(10 ** 6)
input = sys.stdin.readline
N, M = [int(x) for x in input().strip().split()]
L = [deque([]) for _ in range(N)]

for m in range(M):
    l, r, d = [int(x) for x in input().strip().split()]
    L[l-1].append((r, d))
    L[r-1].append((l, -d))

D = [0] * N
F = [False] * N

def dfs(n, d):
    # print('n = {}, d = {}'.format(n, d))
    if F[n-1]:
        if D[n-1] == d:
            # print('  already passed')
            return True
        else:
            # print('  already passed, but wrong')
            return False
    F[n-1] = True
    D[n-1] = d
    for nn, d_ in L[n-1]:
        # print('  dfs({}, {})'.format(nn, d+d_))
        if not dfs(nn, d+d_):
            return False
    else:
        return True

for n in range(1, N+1):
    if not dfs(n, D[n-1]):
        print('No')
        exit()
else:
    print('Yes')