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>

#define LL long long
#define VI vector<int>
#define VB vector<bool>
#define VL vector<long long>
#define FOR(i,a,b) for(int i= (a); i<((int)b); ++i)
#define RFOR(i,a) for(int i=(a); i >= 0; --i)
#define FOE(i,a) for(auto i : a)
#define ALL(c) (c).begin(), (c).end()
#define RALL(c) (c).rbegin(), (c).rend()
#define DUMP(x)  cerr << #x << " = " << (x) << endl;
#define SUM(x) std::accumulate(ALL(x), 0LL)
#define MIN(v) *std::min_element(v.begin(), v.end())
#define MAX(v) *std::max_element(v.begin(), v.end())
#define EXIST(v,x) (std::find(v.begin(), v.end(), x) != v.end())
#define BIT(n) (1LL<<(n))
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end());
#define EPS 1e-14

const std::string YES = "YES";
const std::string Yes = "Yes";
const std::string NO = "NO";
const std::string No = "No";

using namespace std;

template <class T> int find_index(T x, const std::vector<T> &v) {
    auto it = std::find(v.begin(), v.end(), x);
    if (it != v.end()) { return (int)std::distance(v.begin(), it); }
    return -1;
}
template<class T> T ceil(T a, T b) { return (a + b - 1) / b; }

template <class T> void print(T x) {
    std::cout << x << std::endl;
}

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

    vector<vector<pair<int, int>>> graph(N);
    FOR(i, 0, M) {
        int l, r, d;
        cin >> l >> r >> d;
        l--; r--;
        graph[l].emplace_back(make_pair(r, d));
        graph[r].emplace_back(make_pair(l, -d));
    }

    vector<bool> used_node(N, false);
    vector<int> dist(N);

    FOR(i, 0, N) {
        if (used_node[i]) { continue; }

        dist[i] = 0;
        used_node[i] = true;

        queue<int> que;
        que.push(i);
        while (not que.empty()) {
            auto node1 = que.front(); que.pop();


            FOE(node, graph[node1]) {
                int node2 = node.first;
                int d = node.second;

                if (used_node[node2]) {
                    if (dist[node2] != dist[node1] + d) {
                        print(No);
                        return 0;
                    }
                }
                else {
                    used_node[node2] = true;
                    dist[node2] = dist[node1] + d;
                    que.push(node2);
                }
            }
        }
    }

    print(Yes);

}