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 (7.3.0) (judge verdict: Wrong Answer)

import sys
from collections import deque

read = sys.stdin.read
readline = sys.stdin.readline
readlines = sys.stdin.readlines
sys.setrecursionlimit(10 ** 9)
INF = 1 << 60
MOD = 1000000007


class UnionFindWeighted:
    # Reference: https://note.nkmk.me/python-union-find/
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n
        self.weight = [0] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            root = self.find(self.parents[x])
            self.weight[x] += self.weight[self.parents[x]]
            self.parents[x] = root
            return root

    def union(self, x, y, w):
        rx = self.find(x)
        ry = self.find(y)

        if self.parents[rx] > self.parents[ry]:
            rx, ry, w = ry, rx, -w

        self.parents[rx] += self.parents[ry]
        self.parents[ry] = rx
        self.weight[ry] = self.weight[x] - self.weight[y] - w

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def diff(self, x, y):
        return self.weight[x] - self.weight[y]


def main():
    N, M, *LRD = map(int, read().split())

    uf = UnionFindWeighted(N)
    for l, r, d in zip(*[iter(LRD)] * 3):
        l -= 1
        r -= 1
        if uf.same(l, r) and uf.diff(l, r) != d:
            print('No')
            return
        else:
            uf.union(l, r, d)

    print('Yes')
    return


if __name__ == '__main__':
    main()