Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

3 3
1 2 1
2 3 1
1 3 2

Sample Output 1

Yes

Some possible sets of values (x_1, x_2, x_3) are (0, 1, 2) and (101, 102, 103).

Sample Input 2

3 3
1 2 1
2 3 1
1 3 5

Sample Output 2

No

If the first two pieces of information are correct, x_3 - x_1 = 2 holds, which is contradictory to the last piece of information.

Sample Input 3

4 3
2 1 1
2 3 5
3 4 2

Sample Output 3

Yes

Sample Input 4

10 3
8 7 100
7 9 100
9 8 100

Sample Output 4

No

Sample Input 5

100 0

Sample Output 5

Yes

Solution in PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

def main():
    import sys
    input = sys.stdin.readline
    sys.setrecursionlimit(10**7)
    from collections import Counter, deque
    #from collections import defaultdict
    from itertools import combinations, permutations, accumulate, groupby
    #from itertools import product
    from bisect import bisect_left,bisect_right
    from heapq import heapify, heappop, heappush
    from math import floor, ceil
    #from operator import itemgetter

    inf = 10**17
    #mod = 10**9 + 7

    n,m = map(int, input().split())

    #[行先, 距離]
    adj = [[] for _ in range(n)] #頂点数, 場合によって変える
    for _ in range(m):
        l,r,d = map(int, input().split())
        adj[l-1].append([r-1, d])
        adj[r-1].append([l-1, -d])

    x = [inf]*n

    def dfs(v):
        if x[v]==inf:
            x[v] = 0
        d = deque([v])
        while d:
            v  = d.popleft()
            for nv, cost in adj[v]:
                if x[nv] == inf:
                    x[nv] = x[v] + cost
                else:
                    if x[nv] - x[v] != cost:
                        return False
                d.append(nv)
        return True


    for v in range(n):
        if not dfs(v):
            print('No')
            exit()
    print('Yes')


if __name__ == '__main__':
    main()