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>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
#define REP(i, n) for(int i=0; i<n; i++)
const ll INF = 1e10;

typedef struct{
    int to;
    int cost;
}VE;

int N, M;
vector<VE> L[100100];
ll C[100100];

bool dfs(int v, int n){
    C[v] = n;
    for(int i = 0; i < L[v].size(); i++){
        VE ve = L[v][i];
        int next = ve.to;
        int cost = ve.cost;
        if(C[next] != INF){
            if(C[next] != C[v]+cost) {
                return false;
            }
        }
        else{
            bool valid = dfs(next, n+cost);
            if(!valid) return false;
        }
    }
    return true;
}

int main(){
    cin >> N >> M;
    REP(i,M){
        ll l, r, d;
        cin >> l >> r >> d;
        l--, r--;
        L[l].push_back((VE){r, d});
        L[r].push_back((VE){l, -d});
    }

    fill(C, C+N, INF);
    REP(i,N){
        if(C[i] != INF) continue;
        bool valid = dfs(i, 0);
        if(!valid){
            cout << "No" << endl;
            return 0;
        }
    }

    cout << "Yes" << endl;
    return 0;

}