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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

bool dfs(long v, long pos, long *x, bool *vis, vector<vector<pair<long, long>>> &g){
    x[v] = pos;
    vis[v] = true;
    bool ans = true;
    for(auto u : g[v]){
        long to = u.first;
        long dist = u.second;
        if(vis[to]){
            if(x[to] != pos + dist) return false;
        }
        else{
            ans = ans & dfs(to, pos + dist, x, vis, g);
        }
    }
    return ans;
}

int main(){
    long N, M;
    cin >> N >> M;
    long r, l, d, x[N]; 
    bool vis[N]; fill(vis, vis + N, false);
    vector<vector<pair<long, long>>> G(N);
    for(long i=0; i<M; i++){
        cin >> r >> l >> d;
        r --; l --;
        G[r].push_back({l, d});
        G[l].push_back({r, -d});
    }
    bool ans = true;  // ture means satisfaction
    for(long i=0; i<N; i++){
        if(!vis[i]){
            ans = dfs(i, 0, x, vis, G);
            if(!ans) {
                cout << "No" << endl;
                return 0;
            }
        }
    }
    cout << "Yes" << endl;
    return 0;
}