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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <assert.h>

using namespace std;

struct Edge
{
  int to, cost;
};

int main()
{
  int N, M;
  cin >> N >> M;
  vector<Edge> edge[N];
  for (int i = 0; i < M; i++)
  {
    int L, R, D;
    cin >> L >> R >> D;
    L--;
    R--;
    edge[L].push_back({R, D});
    edge[R].push_back({L, -D});
  }
  bool ok = true;
  vector<long long> cost(N, 1e18);
  for (int i = 0; i < N; i++)
    if (cost[i] == 1e18)
    {
      queue<int> que;
      que.push(i);
      cost[i] = 0;
      while (que.size())
      {
        int now = que.front();
        que.pop();
        for (Edge t : edge[now])
          if (cost[t.to] == 1e18)
          {
            cost[t.to] = cost[now] + t.cost;
            que.push(t.to);
          }
          else if (cost[t.to] != cost[now] + t.cost)
            ok = false;
      }
    }
  if (ok)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;

  return 0;
}