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)

/*
	Swamy Saranam

	Date	: 28/01/2018 18:02:54
	Author	: Krishna Mohan A M
	Problem	:
	Status	:
*/
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;

typedef long long ll;
typedef long double ld;
typedef pair<int,int>	pii;
typedef pair<ld,ld>	pdd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef pair<ll,ll> pll;

#define FASTIO                ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define FOR(i,a,b)            for(int i=(a);i<(b);i++)
#define REP(i,n)              FOR(i,0,n)
#define SORT(v)               sort((v).begin(),(v).end())
#define UN(v)                 SORT(v),(v).erase(unique((v).begin(),(v).end()),(v).end())
#define CL(a,b)               memset(a,b,sizeof a)
#define PB                    push_back
#define F                     first
#define S                     second

const int MOD = 1000000007;
const int N = 2e5+5;

ll n, m, dist[N], l, r, d;
vector<pll> graph[N];
bool isPos;

void dfs(int u, int p, ll len){
      // cout<<u<<" : "<<p<<" : "<<len<<"\n";
      dist[u] = len;
      for(pll v : graph[u]){
            if(v.F==p)
                  continue;
            if(dist[v.F] != -1){
                  ll ld = abs(dist[u] - dist[v.F]);
                  // cout<<u<<" - "<<v.F<<" - "<<ld<<" - "<<v.S<<"\n";
                  if(ld != v.S){
                        isPos = false;
                        // cout<<"$"<<isPos<<"\n";
                        return;
                  }
            }
            if(dist[v.F]==-1)
                  dfs(v.F, u, len+v.S);
            if(!isPos)
                  return;
      }
      // cout<<u<<" : "<<isPos<<"\n";
}

int main()
{
      FASTIO
      CL(dist, -1);
      cin>>n>>m;
      REP(i, m){
            cin>>l>>r>>d;
            graph[l].PB({r, d});
            graph[r].PB({l, d});
      }
      isPos = true;
      FOR(i, 1, n+1){
            if(dist[i]==-1){
                  // cout<<"CC: "<<i<<"\n";
                  dist[i] = 0;
                  dfs(i, -1, 0);
            }
            // cout<<"#"<<isPos<<"\n";
            if(!isPos)
                  break;
      }
      // cout<<isPos<<"\n";
      if(!isPos)
            cout<<"No\n";
      else
            cout<<"Yes\n";
      return 0;
}