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 <cmath>
#include <string>
#include <algorithm>
#include <set>
#include <vector>
#include <map>
#include <list>
#include <queue>
#include <stack>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#define REP(i, n) for(i=0; i<n; i++)
#define REPR(i, n) for(i=n; i>=0; i--)
#define FOR(i, m, n) for(i=m; i<n; i++)
#define max(a, b) ((a)>(b) ? (a):(b))
#define min(a, b) ((a)<(b) ? (a):(b))
#define INF 10000000000
#define MOD 1000000007LL
#define MAX 100005
#define NIL -1
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<double, double> doup;

ll N, M;
ll x[100001];
vector<P> G[100000];
ll ans;

void bfs(ll s) {
    ll i, u;
    P v;
    queue<ll> q;
    x[s]=0;
    q.push(s);
    
    while(q.size()) {
        u=q.front(); q.pop();
        
        for(i=0; i<G[u].size(); i++) {
            v=G[u][i];
            if(x[v.first]==INF) {
                x[v.first]=x[u]+v.second;
                q.push(v.first);
            }else if(x[v.first]!=x[u]+v.second) {
                ans=0;
                return;
            }
            
        }
        
    }
    
}




int main(void) {
    ll l, r, d, i;
    cin >> N >> M;
    ans=1;
    for(i=0; i<M; i++) {
        cin >> l >> r >> d;
        l--; r--;
        G[l].push_back(P(r, d));
        G[r].push_back(P(l, -d));
    }
    
    for(i=0; i<N; i++) {
        x[i]=INF;
    }
    
    
    for(i=0; i<N; i++) {
        if(x[i]!=INF) continue;
        bfs(i);
        if(ans==0) {
            cout << "No" << endl;
            return 0;
        }
    }
    
    cout << "Yes" << endl;
}