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: Runtime Error)

// review

#include <iostream>
#include <vector>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
const ll INF = 1e16;
const int MAXN = 110000;
const int MAXM = 210000;


ll l[MAXM], r[MAXM], d[MAXM];
ll x[MAXN];
vector<P> G[MAXN];
ll n, m;

bool dfs(int s) {
  for (int i=0;i<G[s].size();i++) {
    P p = G[s][i];
    if (x[p.first] == INF) {
      x[p.first] = x[s] + p.second;
      if (!dfs(p.first)) {
        return false;
      }
    }
    if (x[p.first] != x[s] + p.second) {
      return false;
    }
  }
  return true;
}

void solve() {
  for (int i=0;i<m;i++) {
    G[r[i] - 1].push_back(P(l[i] - 1, d[i]));
    G[l[i] - 1].push_back(P(r[i] - 1, -d[i]));
  }
  for (int i=0;i<n;i++) {
    x[i] = INF;
  }
  bool success = true;
  for (int i=0;i<n;i++) {
    if (x[i] == INF) {
      if (!dfs(i)) {
        success = false;
        break;
      }
    }
  }
  cout << (success ? "Yes" : "No") << endl;
}

int main() {
  cin >> n >> m;
  for (int i=0;i<m;i++) {
    cin >> l[i] >> r[i] >> d[i];
  }
  solve();
}