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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(n);++i)
#define rrep(i,n) for(int i=1;i<(n);++i)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define maxs(a, b) a = max(a, b)
#define mins(a, b) a = min(a, b)
using namespace std;
template<class T> inline istream& operator >> (istream & c, vector<T> & A) {rep(i, A.size()) c >> A[i];return c;}
typedef long long ll;
typedef pair<int, int> P;
const ll linf = (1ll << 61);
const int inf = 1001001001;
const int mod = 1000000007;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n, m;
	cin >> n >> m;
	vector<vector<P>> g(n);
	rep(i, m) {
		int l, r, d;
		cin >> l >> r >> d;
		l--; r--;
		g[l].push_back(P(r, d));
		g[r].push_back(P(l, -d));
	}
	vector<int> memo(n, -inf);
	bool ok = true;
	function<void(int, int)> Dfs = [&](int v, int now) {
		if (memo[v] != -inf) {
			if (memo[v] != now) ok = false;
			return;
		}
		memo[v] = now;
		for (auto c : g[v]) {
			Dfs(c.first, memo[v] + c.second);
		}
		return;
	};
	rep(i, n) {
		if (memo[i] == -inf) Dfs(i, 0);
	}
	if (ok) cout << "Yes" << endl;
	else cout << "No" << endl;
	return 0;
}