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>
using namespace std;

#define ll long long
#define vvi vector<vector<int>>
#define vec vector
#define pq priority_queue
#define all(v) (v).begin(), (v).end()
#define rep(i, n, m) for (int (i) = (n); (i) < (m); (i)++)
#define debug(x) cerr << #x << ": " << x << endl;
#define debug2(x, y) cerr<<"("<<#x<<", "<<#y<<") = "<<"("<<x<<", "<<y<<")"<<endl;
#define debugB(x, y) cerr<<#x<<": "<<bitset<y>(x) << endl;
#define line() cerr << "---------------" << endl;

struct Edge {
   int to, cost;
};
ll INF = 1e18;
int N, M;
vec<vec<Edge>> G;

void dfs(int pos, vec<ll> &dist) {
   for (Edge e : G[pos]) {
      if (dist[e.to] == INF) {
         dist[e.to] = dist[pos] + e.cost;
         dfs(e.to, dist);
      }
   }
}

int main() {
   cin >> N >> M;
   G = vec<vec<Edge>>(N);
   rep (i, 0, M) {
      int l, r, d;
      cin >> l >> r >> d;
      l--, r--;
      G[l].push_back(Edge{r, d});
      G[r].push_back(Edge{l, -d});
   }

   vec<ll> dist(N, INF);
   rep (i, 0, N) {
      if (dist[i] == INF) {
         dist[i] = 0;
         dfs(i, dist);
      }
   }

   rep (i, 0, N) {
      for (Edge e : G[i]) {
         if (dist[e.to] - dist[i] != e.cost) {
            puts("No");
            return 0;
         }
      }
   }
   puts("Yes");

   return 0;
}