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 <iostream>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <utility>
#include <climits>

typedef long vertex;
typedef long dist;

using namespace std;

vector<bool> data_exists;
vector<long> xs;

long n;
long m;

vector<vertex> ls;
vector<vertex> rs;
vector<dist> ds;

map<vertex, vector<pair<vertex,dist>>> adj_map;

bool dfs(vertex v) {
    for (auto&& pair : adj_map[v]) {
        vertex w = pair.first;
        dist d = pair.second;
        if (data_exists[w]) {
            if (xs[w] != xs[v] + d) return false;
        } else {
            data_exists[w] = true;
            xs[w] = xs[v] + d;
            if (!dfs(w)) return false;
        }
    }
    return true;
}

bool solve() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    cin >> n >> m;
    data_exists.resize((unsigned long)n, false);
    xs.resize((unsigned long)n);

    for (int i = 0; i < m; i++) {
        vertex l, r;
        dist d;
        cin >> l >> r >> d;
        l--;
        r--;
        adj_map[l].emplace_back(make_pair(r, d));
        adj_map[r].emplace_back(make_pair(l, -d));
    }

    for (vertex v = 0; v < n; v++) {
        if (!data_exists[v]) {
            xs[v] = 0;
            if (!dfs(v)) return false;
        }
    }
    return true;
}

int main() {
    cout << (solve() ? "Yes" : "No");
    return 0;
}