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)

#include <iostream>
#include <vector>
#include <queue>
#include <map>
#include <string>
#include <algorithm>
#include <cmath>
using namespace std;
#define dump(a) cout << (a) << endl;
typedef long long ll;

bool node[100010];
ll location[100010];
vector<pair<int, int>> graph[100005];

bool dfs( int x , int cost ) {
    if( node[x] ) return location[x] != cost;
    node[x] = true;
    location[x] = cost;
    for( int i = 0; i < (int)graph[x].size(); i++ ) {
        return dfs( graph[x][i].first, cost+graph[x][i].second );
    }
    return false;
}   

int main() {
    int n, m;
    cin >> n >> m;
    for( int i = 0; i < m; i++ ) {
        int a, b, c;
        cin >> a >> b >> c;
        graph[a].push_back(make_pair(b, c));
        graph[b].push_back(make_pair(a, -c));
    }
    for( int i = 1; i <= n; i++ ) node[i] = false;
    bool res = true;
    for( int i = 1; i <= n; i++ ) {
        if( node[i] ) continue;
        if( dfs(i, 0) ) res = false;
    }
    dump( res ? "Yes" : "No" );
    return 0;  
}