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:
#!/usr/bin/env python3
import sys
from collections import defaultdict, deque

YES = "Yes"  # type: str
NO = "No"  # type: str

def solve(N: int, M: int, adj):
    visited = [False] * N
    pos = [None] * N

    for i, vis in enumerate(visited):
        if vis:
            continue
        pos[i] = 0
        visited[i] = True
        Q = deque([i], M)
        while len(Q) > 0:
            u = Q.popleft()
            for v, d in adj[u]:
                if pos[v] is None:
                    pos[v] = pos[u] + d
                elif pos[v] != pos[u] + d:
                    print(NO)
                    return

                if visited[v]:
                    continue
                Q.append(v)
                visited[v] = True
    print(YES)
    return


# Generated by 1.1.4 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    N = int(next(tokens))  # type: int
    M = int(next(tokens))  # type: int
    adj = [set() for _ in range(N)]
    for i in range(M):
        l = int(next(tokens)) - 1
        r = int(next(tokens)) - 1
        d = int(next(tokens))
        adj[l].add((r, d))
        adj[r].add((l, -d))
    solve(N, M, adj)

if __name__ == '__main__':
    main()