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: Accepted)

import sys
input = sys.stdin.readline

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

def Find(x, par, diff_weight):
  if par[x] < 0:
    return x
  else:
    # 経路圧縮
    px = Find(par[x], par, diff_weight)
    diff_weight[x] += diff_weight[par[x]]
    par[x] = px
    return px

def Unite(x, y, par, rank, diff_weight, w):
  w += diff_weight[x] - diff_weight[y]
  x = Find(x, par, diff_weight)
  y = Find(y, par, diff_weight)
  
  if x != y:
    # rankの低い方を高い方につなげる
    if rank[x] < rank[y]:
      x, y =  y, x
      w = -w
    if rank[x] == rank[y]:
        rank[x] += 1
    par[x] += par[y]
    par[y] = x
    diff_weight[y] = w

def Same(x, y, par, diff_weight):
  return Find(x, par,diff_weight) == Find(y, par, diff_weight)
 
def Size(x, par):
  return -par[Find(x)]

def Weight(x, par, diff_weight):
  Find(x, par, diff_weight)
  return diff_weight[x]

def Diff(x, y, par, diff_weight):
  return Weight(y, par, diff_weight)  - Weight(x, par, diff_weight)

par = [-1]*n
rank = [0]*n
diff_weight = [0]*n
for i in range(m):
  l, r, d = map(int, input().split())
  l, r = l-1 ,r-1
  if Same(l, r, par, diff_weight) == False:
    Unite(l, r, par, rank, diff_weight, d)
  else:
    if d != Diff(l, r, par, diff_weight):
      print('No')
      exit()
print('Yes')