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;

class UnionFind{
    vector<int> par;
    vector<long long> wei;
    int groupCount;

public:
    UnionFind(int n = 0){
        init(n);
    }

    void init(int n = 0){
        par.resize(n);
        fill(par.begin(), par.end(), -1);

        wei.resize(n);
        fill(wei.begin(), wei.end(), 0LL);

        groupCount = n;

        return;
    }

    int root(int x){
        if(par[x] < 0){
            return x;
        }
        int r = root(par[x]);
        wei[x] += wei[par[x]];
        return par[x] = r;
    }

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

    int size(int x){
        return -par[root(x)];
    }

    long long weight(int x){
        root(x);
        return wei[x];
    }

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

    bool unite(int x, int y, long long w = 0){
        w += weight(x) - weight(y);

        x = root(x);
        y = root(y);

        if(x == y){
            return false;
        }

        if(par[y] < par[x]){
            swap(x, y);
            w = -w;
        }

        par[x] += par[y];
        par[y] = x;
        wei[y] = w;
        --groupCount;

        return true;
    }

    int size(void){
        return groupCount;
    }
};

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n, m;
    cin >> n >> m;

    UnionFind uf(n);
    for(int i = 0 ; i < m ; ++i){
        int l, r;
        long long d;
        cin >> l >> r >> d;
        if(uf.same(--l, --r) && uf.diff(l, r) != d){
            cout << "No" << endl;
            return 0;
        }
        uf.unite(l, r, d);
    }

    cout << "Yes" << endl;

    return 0;
}