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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

#define reps(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   reps(i,0,n)
#define all(x) (x).begin(),(x).end()
#define INF (1000000000)
#define MOD (1000000007)
#define PI (acos(-1))

class UnionFind{
  public:
  //親の番号を格納する 親だった場合-1*集合のサイズ
  vector<int> Parent;
  vector<int> Diff_weight;

  //コンストラクタ 初めはバラバラの状態
  UnionFind(int N){
    Parent = vector<int>(N,-1);//要素数N,要素の値-1で初期化
    Diff_weight = vector<int>(N,0);//親への距離
  }

  //Ａがどのグループに属しているか調べる
  int root(int A){
    if(Parent[A] < 0){
      return A;//Aが親だったらA自身を返す
    }else{
      int tmp = root(Parent[A]);
      Diff_weight[A] += Diff_weight[Parent[A]];
      return Parent[A] = tmp;//親に繋ぎなおす
    }
  }

  int weight(int A){
    root(A);
    return Diff_weight[A];
  }

  //頂点数を調べる
  int size(int A){
    return -Parent[root(A)];//Aの根×-1=集合のサイズ
  }

  //AとBをくっつける
  bool connect(int A, int B, int W){
    W += Diff_weight[A]; W -= Diff_weight[B];
    //Aの根とBの根をくっつける
    A = root(A);
    B = root(B);
    if(A==B)return false;//既に同じグループならそのままでfalseを返す

    //Aのサイズ>=Bのサイズとする  
    if(size(A)<size(B))swap(A,B);

    Parent[A] += Parent[B];//AのサイズにBのサイズを足す
    Parent[B] = A;//Bの親をAに変更する

    Diff_weight[B] = W;
    return true;
  }

  //AとBが同じグループかどうか調べる
  bool same(int A,int B){
    if(root(A)==root(B))return true;
    else return false;
  }

};

int main(){
  int N,M;
  cin >> N >> M;

  UnionFind Uni(N);

  int L,R,D;
  rep(i,M){
    cin >> L >> R >> D;
    L--;R--;
    if(Uni.same(L,R)){
      if(Uni.weight(R)-Uni.weight(L)!=D){
        cout << "No" << endl;
        return 0;
      }
    }
    Uni.connect(L,R,D);
  }
  cout << "Yes" << endl;
}