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: Runtime Error)

#include <algorithm>
#include <climits>
#include <iostream>
#include <queue>
#include <stack>
#include <utility>
#include <vector>

#define REP(i, n) for (int i = 0; i < n; i++)
#define FOR(i, m, n) for (int i = m; i < n; i++)
#define ALL(a) (a).begin(), (a).end()

#define INF 1000000000

using namespace std;

int main() {
    int N, M;
    cin >> N >> M;

    vector<vector<pair<int, int>>> g(N + 100);

    REP(i, M) {
        int s, t;
        cin >> s >> t;

        s--;
        t--;

        int w;
        cin >> w;

        g[s].emplace_back(t, w);
        g[t].emplace_back(s, -w);
    }

    vector<bool> visited(N, false);
    vector<long long> dist(N, INF);

    REP(i, N) {
        if (visited[i] == false) {
            queue<pair<int, long long>> q;
            q.emplace(i, 0);

            while (!q.empty()) {
                auto& p = q.front();
                q.pop();

                for (auto& e : g[p.first]) {
                    if (visited[e.first]) {
                        if (dist[e.first] != p.second + e.second) {
                            cout << "No" << endl;
                            return 0;
                        }
                    } else {
                        int next_dist = p.second + e.second;
                        dist[e.first] = next_dist;
                        visited[e.first] = true;
                        q.emplace(e.first, next_dist);
                    }
                }
            }
        }
    }

    cout << "Yes" << endl;

    return 0;
}