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;

struct st {
    int L, R, D;
};

int main() {
    int N, M;
    cin >> N >> M;
    if (M == 0) {
        cout << "Yes" << endl;
        return 0;
    }
    vector<st> S(M);
    vector<vector<pair<int, int>>> mp(N + 1);
    for (int i = 0; i < M; i++) {
        cin >> S[i].L >> S[i].R >> S[i].D;
        mp[S[i].L].push_back({ S[i].R, S[i].D });
        mp[S[i].R].push_back({ S[i].L, -S[i].D });

        if (S[i].L > S[i].R) {
            swap(S[i].L, S[i].R);
            S[i].D *= -1;
        }
        //mp[S[i].L].push_back({ S[i].R, S[i].D });
    }

    sort(S.begin(), S.end(), [&](const st& lsh, const st& rhs) {
        return (lsh.L < rhs.L);
    });

    vector<long long> pos(N + 1, INT_MIN);

    stack<int> sta;
    sta.push(S[0].L);
    pos[S[0].L] = 0;
    while (!sta.empty()) {
        int target = sta.top();
        sta.pop();

        for (auto p : mp[target]) {
            if (pos[p.first] != INT_MIN) {
                if (pos[p.first] - pos[target] != p.second) {
                    cout << "No" << endl;
                    return 0;
                }
            } else {
                pos[p.first] = pos[target] + p.second;
                sta.push(p.first);
            }
        }
    }

    cout << "Yes" << endl;
}