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 <cstdio>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>

using namespace std;

typedef long long ll;
typedef pair<int, int> P;

#define rep(i, n) for (int i = 0; i < (n); i++)
#define repr(i, n) for (int i = n; i >= 0; i--)
#define FOR(i, m, n) for (int i = m; i < n; i++)
#define FORR(i, m, n) for (int i = m; i >= n; i--)

int const M_MAX = 200000;

int N, M;

vector<P> G[M_MAX + 1];
map<int, int> used;
map<int, map<int, ll>> costs;

int bfs(int current) {
  queue<int> Q;
  used[current]++;
  Q.push(current);

  while(!Q.empty()) {
    int idx = Q.front();
    Q.pop();

    for (auto p : G[idx]) {
      ll cost = costs[current][p.first];
      if (cost) {
        if (cost != costs[current][idx] + p.second) return 0;
      } else {
        costs[current][p.first] = costs[current][idx] + p.second;
        costs[p.first][current] = -1 * costs[current][p.first];
      }

      if (used[p.first]) continue;
      used[p.first]++;
      Q.push(p.first);
    }
  }
  return 1;
}

int main() {
  scanf("%d %d", &N, &M);

  rep(i, M) {
    int l, r, d;
    scanf("%d %d %d", &l, &r, &d);

    G[l].push_back(P(r, d));
    G[r].push_back(P(l, -1 * d));
  }

  rep(i, N + 1) {
    if(!bfs(i)) {
      printf("No\n");
      return 0;
    }
  }
  printf("Yes\n");
}