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)

import sys
sys.setrecursionlimit(100000)

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

graph = [{} for _ in range(N)]
for l, r, d in LRD:
	l, r = l - 1, r - 1
	graph[l][r] = d
	graph[r][l] = -d

visited = [False] * N
def dfs(v, p, d, depths):
	if visited[v]:
		return True
	visited[v] = True

	if depths[v] != -1:
		return True
	depths[v] = d

	for n in graph[v]:
		if n == p:
			continue
		n_depth = d + graph[v][n]
		if depths[n] != -1 and depths[n] != n_depth:
			return False
		result = dfs(n, v, n_depth, depths)
		if not result:
			return False
	return True

for n in range(N):
	if not dfs(n, -1, 0, [-1] * N):
		print("No")
		break
else:
	print("Yes")