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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <cmath>
#include <iomanip>
#include <stack>
#include <queue>
#include <numeric>
#include <map>
#include <unordered_map>
#include <set>
#include <fstream>
#include <chrono>
#include <random>
#include <bitset>
#define rep(i,n) for(int i=0;i<(n);i++)
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define sz(x) ((int)(x).size())
#define pb push_back
#define mod 1000000007
using ll = long long;
using namespace std;
ll gcd(ll a, ll b) {return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) {return a/gcd(a,b)*b;}

int main(){
  ll N, M; cin >> N >> M;
  if(M<=2) {
    cout << "Yes" << endl;
    return 0;
  }
  vector<pair<pair<ll,ll>,ll>> LRD(M);
  rep(i,M) {
    cin >> LRD[i].first.first >> LRD[i].first.second >> LRD[i].second;
    LRD[i].first.first--;
    LRD[i].first.second--;
  }
  vector<vector<ll>> g(N);
  vector<vector<ll>> c(N);
  rep(i,M){
    ll l = LRD[i].first.first, r = LRD[i].first.second, d = LRD[i].second;
    g[l].pb(r);
    c[l].pb(d);
    g[r].pb(l);
    c[r].pb(-d);
  }
  vector<ll> pos(N,1e18);
  vector<bool> chk(N,false);
  queue<ll> q;
  rep(m,M){
    if(pos[LRD[m].first.first]==1e18){
      pos[LRD[m].first.first] = 0;
      q.push(LRD[m].first.first);
    }
    while(!q.empty()){
      ll l = q.front(); q.pop();
      if(chk[l]) continue;
      chk[l] = true;
      rep(i,sz(g[l])){
        ll r = g[l][i];
        if(pos[r]!=1e18){
          if(pos[r]-pos[l]!=c[l][i]){
            cout << "No" << endl;
            return 0;
          }
        }
        else{
          pos[r] = pos[l]+c[l][i];
          q.push(r);
        }
      }
    }
  }
  cout << "Yes" << endl;
  return 0;
}