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>
#include<unordered_set>
#include<unordered_map>
#include <iostream>
#include <string>
#include <cmath>
using namespace std;
#define ll long long
#define rep(i, n) for (ll i = 0; i < (n); i++)
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define FORR(i,a,b)for(ll i=(a);i<=(b);i++)
#define repR(i,n) for(ll i=n;i>=0;i--)
#define all(v)(v).begin(),(v).end()
#define rall(v)(v).rbegin(),(v).rend()
#define F first
#define S second
#define pb push_back
#define pu push
#define COUT(x) cout<<(x)<<endl
#define PQ priority_queue<ll>
#define PQR priority_queue<ll,vector<ll>,greater<ll>>
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define mp make_pair
#define maxs(x,y) (x = max(x,y))
#define mins(x,y) (x = min(x,y))
#define sz(x) (int)(x).size()
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
const ll MOD = 1000000007LL;
const ll INF = 1LL << 60;
using vll = vector<ll>;
using vb = vector<bool>;
using vvb = vector<vb>;
using vvll = vector<vll>;
using vstr = vector<string>;
using pll = pair<ll, ll>;
using vc = vector<char>;
using vvc = vector<vc>;
ll dx[4]={0,1,0,-1};
ll dy[4]={1,0,-1,0};
int main(){
  ll n,m;
  cin>>n>>m;
  vector<vector<pll>> ed(n+1,vector<pll>(0));
  rep(i,m){
    ll l,r,d;
    cin>>l>>r>>d;
    ed[r].pb(pll(l,-d));
    ed[l].pb(pll(r,d));
  }
  queue<ll> que;
  vll dist(n+1,INF);
  for(int i=1;i<=n;i++){
    if(dist[i]==INF){
      dist[i]=0;
      que.pu(i);
      while(sz(que)){
        ll from=que.front();
        que.pop();
        for(pll p:ed[from]){
          if(dist[p.F]==INF){
            dist[p.F]=dist[from]+p.S;
            que.pu(p.F);
          }
          else if(dist[p.F]!=dist[from]+p.S){
            COUT("No");
            return 0;
          }
        }
      }
    }
  }
  COUT("Yes");
}