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 long long ll;

int n, m;
vector<pair<int, int>> e[100005];

ll x[1000005];
bool vis[1000005];

bool ok = true;

void solve(int v) {
	vis[v] = true;
	for (auto p : e[v]) {
		int y = p.first;
		int d = p.second;

		if (vis[y]) {
			if (x[v] + d != x[y]) {
				ok = false;
			}
		} else {
			x[y] = x[v] + d;
			solve(y);
		}
		
	}
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	cerr.tie(nullptr);

	cin >> n >> m;
	for (int i=1; i<=m; i++) {
		int l, r, d;
		cin >> l >> r >> d;
		e[l].push_back({r, d});
		e[r].push_back({l, -d});
	}

	for (int i=1; i<=n; i++) {
		if (!vis[i]) {
			solve(i);
		}
	}

	cout << (ok ? "Yes" : "No") << '\n';
}