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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define REP(i, n) for(int i = 0; i < (n); ++i)
#define REPI(itr, v) for(auto itr = (v).begin(); itr != (v).end(); ++itr)

using namespace std;

typedef struct{
  int to,cost;
} edge;


bool is_line(vector<vector<edge>> G){

  int N = G.size();
  vector<bool> is_visit(N, false);

  REP(i, N){
    if(is_visit[i]){
      continue;
    }
    
    queue<int> que;
    vector<int> x(N, 0);

    que.push(i);
    while(!que.empty()){
      int j = que.front();
      que.pop();
      REPI(itr, G[j]){
	int k = itr->to;
	int x_new = x[j] + itr->cost;
	if(is_visit[k]){
	  if(x[k] != x_new){
	    return false;
	  }
	}else{
	  que.push(k);
	  is_visit[k] = true;
	  x[k] = x_new;
	}
      }
    }
  }

  return true;
}

int main(void){

  int N, M;
  cin >> N >> M;

  vector<vector<edge>> G(N);
  REP(i, M){
    int L, R, D;
    cin >> L >> R >> D;
    edge e1, e2;
    e1.to = R-1, e1.cost = D;
    e2.to = L-1, e2.cost = -1 * D;
    G[L-1].push_back(e1);
    G[R-1].push_back(e2);
  }

  cout << (is_line(G) ? "Yes" : "No") << endl;

  return 0;
}