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: Runtime Error)

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=int(a);i<int(b);++i)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int INF = (1LL << 30) - 1;
int MOD = 1e9+7;
vector<ll> x;
vector<P> V[100000];
bool dfs(int t){
    bool res = false;
    for(auto p:V[t]){
        if(x[p.first]){
            if(x[t] + p.second != x[p.first]){
                //cout << t << " " << x[t] << " " << p.second << " " << x[p.first] << endl;
                return true;
            }
        }else{
            x[p.first] = x[t] + p.second;
            res |= dfs(p.first);
        }
    }
    return res;
}
main(){
    int N,M;
    cin >> N >> M;
    x = vector<ll>(N);
    rep(i,0,M){
        int a,b,c;
        cin >> a >> b >> c;
        a--,b--;
        V[a].push_back(P(b,c));
        V[b].push_back(P(a,-c));
    }
    rep(i,0,N){
        if(!x[i] && dfs(i)){
            cout << "No" << endl;
            return 0;
        }
    }
    cout << "Yes" << endl;
}