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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pp;
const int INF = 1e9;
const int MOD = 1000000007;
#define rep(i,n) for(int i=0;i<n;i++)

struct potential_Unionfind{
    vector<int> parents;
    vector<int> weights;
    int m;
    potential_Unionfind(int n):parents(n,-1){
        m = n;
        weights.resize(n);
    }

    int find(int x){
        if(parents[x] < 0) return x;
        else{
            parents[x] = find(parents[x]);
            weights[x] += weights[parents[x]];
            return parents[x];
        }
    }

    //weight[y] - weights[x] = wとなるようにuniteする。
    bool unite(int x,int y,int w){
        w += weights[x];
        w -= weights[y];
        x = find(x);
        y = find(y);
        if(x == y) return false;//もうすでに繋がれている頂点ならfalseを返す
        if(parents[x] > parents[y]) {swap(x,y); w = -w;}
        parents[x] += parents[y];
        parents[y] = x;
        weights[y] = w;
        return true;
    }

    bool same(int x,int y){
        return find(x) == find(y);
    }

    //weightsは各頂点から根までの距離
    int weight(int x){
        find(x);
        return weights[x];
    }

    int diff(int x,int y){
        return weight(y) - weight(x);
    }


};

int main(){
    int n,m;
    cin >> n >> m;
    potential_Unionfind uf(n);
    rep(i,m){
        int a,b,w;
        cin >> a >> b >> w;
        --a;--b;
        if(!uf.unite(a,b,w)){
            if(uf.diff(a,b) != w){
                cout << "No" << endl;
                return 0;
            }
        }
    }
    
    cout << "Yes" << endl;
    return 0;

}