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;
const int nmax=100005;
const int INF=2e9;
struct edge{int to;int cost;};
vector<edge> tree[nmax];
vector<int> d(nmax,-INF);
bool f;
void bfs(int s){
  queue<int> q;
  q.push(s);
  d[s]=0;
  while(!q.empty()){
    int p=q.front();
    q.pop();
    for(auto x:tree[p]){
      int u=x.to,v=x.cost;
      if(d[u]==-INF){
        d[u]=d[p]+v;
        q.push(u);
      }
      else{
        if(d[u]!=d[p]+v){
          f=false;
        }
      }
    }
  }
}
int main(){
  int n,m;cin >> n >> m;
  for(int i=0;i<m;i++){
    int l,r,d;cin >> l >> r >> d;
    l--,r--;
    tree[l].push_back({r,d});
    tree[r].push_back({l,-d});
  }
  f=true;
  for(int i=0;i<n;i++){
    if(d[i]!=-INF){
      continue;
    }
    bfs(i);
    if(!f){
      cout << "No" << endl;
      return 0;
    }
  }
  cout << "Yes" << endl;
}