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;

using ll = long long;

#define REP(i, n) for (int i = 0; i < (n); ++i)
#define SIZE(a) ((int)(a).size())

const int N = 100005;

int n, m;
vector<pair<int,int>> g[N];
int was[N];
int dist[N];

void die() {
	cout << "No\n";
	exit(0);
}

void dfs(int v, int d = 0) {
	was[v] = 1;
	dist[v] = d;
	for (auto& item : g[v]) {
		int to = item.first;
		int c = item.second;
		if (!was[to]) {
			dfs(to, d + c);
		} else {
			if (dist[to] != d + c) {
				die();
			}
		}
	}
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);

	cin >> n >> m;
	REP(i, m) {
		int x, y, c;
		cin >> x >> y >> c;
		--x; --y;
		g[x].emplace_back(y, c);
		g[y].emplace_back(x, -c);
	}
	REP(i, n) if (!was[i]) dfs(i);
	cout << "Yes\n";
	return 0;
}