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<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<cmath>
#include<functional>
#include<map>
#include<cstdlib>
#include<complex>
#include<numeric>
#include<set>
#include<deque>
#include<queue>

using namespace std;
using Int = long long;
Int INF = 1LL << 60;
const Int MOD = 1000000000 + 7;

vector<Int> dist;
vector<vector<pair<Int,Int>>> G;

bool dfs(Int v) {
	
	for (auto e : G[v]) {
		if (dist[e.first] == INF) {//行き先が見探索の場合
			dist[e.first] = dist[v] + e.second;
			//return dfs(e.first);
			dfs(e.first);
		}
		else if (dist[v] + e.second == dist[e.first])continue;//矛盾しない場合
		else return false;//矛盾する場合
	}

	return true;
}

int main() {

	cin.tie(nullptr);
	ios::sync_with_stdio(false);

	Int N, M;
	cin >> N >> M;

	dist.resize(N, INF);
	G.resize(N);

	Int L, R, D;
	for (Int i = 0; i < M; i++) {
		cin >> L >> R >> D;
		L--;
		R--;
		G[L].push_back(make_pair(R, D));
		G[R].push_back(make_pair(L, -D));
	}

	for (Int i = 0; i < N; i++) {
		if (dist[i] == INF) {
			dist[i] = 0;
			if (!dfs(i)) {
				cout << "No" << "\n";
				return 0;
			}
		}
	}

	cout << "Yes" << "\n";
}