Problem: 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`.

Examples

Input

3 3
1 2 1
2 3 1
1 3 2


Output

Yes


Input

3 3
1 2 1
2 3 1
1 3 5


Output

No


Input

4 3
2 1 1
2 3 5
3 4 2


Output

Yes


Input

10 3
8 7 100
7 9 100
9 8 100


Output

No


Input

100 0


Output

Yes

Solution (in CPP):

Solution:
#include "bits/stdc++.h"
using namespace std;
typedef pair<int, int> P;
int x[100000];
vector<P> G[100000];
bool vis[100000];
bool dfs(int v, int now) {
	vis[v] = 1;
	x[v] = now;
	for (auto e : G[v]) {
		int to = e.first;
		int d = e.second;
		if (!vis[to]) {
			if (!dfs(to, now + d)) return false;
		}
		else if (x[v] + d != x[to]) {
			return false;
		}
	}
	return true;
}
int main() {
	int N, M;
	cin >> N >> M;
	for (int i = 0; i < M; i++) {
		int L, R, D;
		cin >> L >> R >> D;
		L--; R--;
		G[L].push_back(P(R, D));
		G[R].push_back(P(L, -D));
	}
	for (int i = 0; i < N; i++) {
		if (!vis[i]) {
			if (!dfs(i, 0)) {
				cout << "No" << endl;
				return 0;
			}
		}
	}
	cout << "Yes" << endl;
}