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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <deque>
using namespace std;

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

class Edge{
public:
  int x;
  int y;
  ll d;
  void set(int st, int go, ll dis){
    x=st;y=go;d=dis;
  }
  bool operator <(const Edge& f){return x<f.x;}
};

int main(){
  int n,m;
  ll loc[100010]={};
  bool ans=true;
  bool used[100010]={};
  Edge e[200010];
  int minp;
  cin>>n>>m;
  minp=n;
  for(int i=0;i<m;i++){
    int x,y;
    ll d;
    cin>>x>>y>>d;
    if(y<x){
      swap(x,y);
      d*=-1;
    }
    minp=min(minp,x);
    e[i].set(x,y,d);
  }
  sort(e,e+m);
  used[minp]=true;
  loc[minp]=0;
  for(int i=0;i<m;i++){
    int x,y;
    ll d;
    x=e[i].x;y=e[i].y;d=e[i].d;
    if(!used[x]){
      loc[x]=0;
      used[x]=true;
    }
    if(used[y]){
      if(loc[x]+d!=loc[y]){
       ans=false;
       cout<<x<<" "<<y<<" "<<loc[x]+d<<" "<<loc[y]<<endl;
       break;
      }
    }else{
      loc[y]=loc[x]+d;
      used[y]=true;
    }
  }
  cout<<(ans?"Yes":"No")<<endl;
  return 0;
}