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;
#define rep(i, n) for(ll i = 0, i##_len = (n); i < i##_len; i++)
#define reps(i, s, n) for(ll i = (s), i##_len = (n); i < i##_len; i++)
#define rrep(i, n) for(ll i = (n) - 1; i >= 0; i--)
#define rreps(i, e, n) for(ll i = (n) - 1; i >= (e); i--)
#define all(x) (x).begin(), (x).end()
#define sz(x) ((ll)(x).size())
#define len(x) ((ll)(x).length())

template<class T>
struct WeightedUnionFind {
private:
    vector<int> par, rank;
    vector<T> diff_weight;
    
public:
    WeightedUnionFind() {}
    WeightedUnionFind(int n, T def = 0) {
        par.resize(n, -1);
        rank.resize(n, 0);
        diff_weight.resize(n, def);
    }
    
    int root(int x) {
        if (par[x] < 0) return x;
        int r = root(par[x]);
        diff_weight[x] += diff_weight[par[x]];
        return par[x] = r;
    }
    
    T weight(int x) {
        root(x);
        return diff_weight[x];
    }
    
    bool is_same(int x, int y) {
        return root(x) == root(y);
    }
    
    bool merge(int x, int y, T w) {
        w += weight(x); w -= weight(y);
        x = root(x); y = root(y);
        if (x == y) return false;
        if (rank[x] < rank[y]) swap(x, y), w = -w;
        if (rank[x] == rank[y]) rank[x]++;
        par[y] = x;
        diff_weight[y] = w;
        return true;
    }
    
    T diff(int x, int y) {
        return weight(y) - weight(x);
    }
    
    int size(int x) {
        return -par[root(x)];
    }
};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    // ifstream in("input.txt");
    // cin.rdbuf(in.rdbuf());
    ll n, m;
    cin >> n >> m;
    WeightedUnionFind<ll> wuf(n);
    rep(i, m) {
        ll l, r, d;
        cin >> l >> r >> d;
        l--; r--;
        if (wuf.is_same(l, r)) {
            if (wuf.diff(l, r) != d) {
                cout << "No" << endl;
                return 0;
            }
        }
        else {
            wuf.merge(l, r, d);
        }
    }
    cout << "Yes" << endl;
    return 0;
}