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 <stdio.h>
#include <string>
#include <memory.h>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <time.h>
#include <assert.h>
#include <cmath>
#include <stack>
#include <string.h>
#include <sstream>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N = 200000;
int n, m;
vector<vector<pair<int, int> > > g;
int loc[N];
bool vis[N];
void DFS(int u, int at) {
	if (vis[u]) {
		if (at != loc[u]) {
			puts("No");
			exit(0);
		}
		return;
	}
	loc[u] = at;
	vis[u] = true;
	for (int i = 0; i < g[u].size(); ++i)
		DFS(g[u][i].first, at + g[u][i].second);
}
int main()
{
	scanf("%d%d", &n, &m);
	g.resize(n);
	for (int i = 0, a, b, c; i < m; ++i) {
		scanf("%d%d%d", &a, &b, &c);
		--a; --b;
		g[a].push_back({ b,c });
		g[b].push_back({ a,-c });
	}
	for (int i = 0; i < g.size(); ++i)
		if (!vis[i]) {
			DFS(i, 0);
		}
	puts("Yes");
	return 0;
}