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 Python (3.4.3) (judge verdict: Accepted)

n, m = list(map(int, input().split()))
par = list(range(n+1))
dis = [0] * (n+1)

def root(x):
  if x == par[x]:
    return x
  r = root(par[x])
  dis[x] += dis[par[x]]
  par[x] = r
  return r

def cost(x):
  # xからxの根までのコスト
  if x == par[x]:
    return 0
  return dis[x] + cost(par[x])

for _ in range(m):
  l, r, d = list(map(int, input().split()))
  # 根が違ってればマージ
  # 同じなら入力されたdistanceがこれまでの入力データからなる
  # 現在の木の構造から計算した数値とあっているか判断
  # rootl, rootr = root(l), root(r)
  # 確かめ用
  rootl = root(l)
  rootr = root(r)
  if rootl != rootr:
    if r != rootr:
      # 直接yにつなげる
      dis[rootl] = d - cost(l)
    else:
      # 根につなげる
      dis[rootl] = d - (cost(l) - cost(r)) # (dis[l] - dis[r])
    # ここで親を更新
    par[rootl] = r
  else:
    if d != cost(l) - cost(r):
      print("No")
      exit()

  """
  ちょっとなにがダメなのかわかんないので確かめる
  if rootl != rootr:
    # lをrにマージする
    if r == rootr:
      dis[rootl] = d - cost(l)
    else:
      dis[rootl] = d - (dis[l] - dis[r]) # (cost(l) - cost(r))
    par[rootl] = r
  else:
    # コストが同じかどうか
    if d == cost(l) - cost(r):
      continue
    else:
      print("No")
      exit()
  """
print("Yes")