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;

const int MAXN = 100000;

vector<pair<int, int>> edges[MAXN];
int dist[MAXN];
bool bio[MAXN];

bool dfs(int i)
{
	for (const auto& edge: edges[i])
	{
		if (!bio[edge.first])
		{
			dist[edge.first] = dist[i] + edge.second;
			bio[edge.first] = true;
			if (dfs(edge.first))
				return true;
		}
		else if (dist[edge.first] != dist[i] + edge.second)
			return true;
	}
	return false;
}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	
	int n, m, i;
	int l, r, d;
	
	cin>>n>>m;
	
	while (m--)
	{
		cin>>l>>r>>d;
		--l, --r;
		
		edges[l].push_back({r, d});
		edges[r].push_back({l, -d});
	}
	
	for (i = 0; i < n; ++i) if (!bio[i])
	{
		dist[i] = 0;
		bio[i] = true;
		if (dfs(i))
		{
			cout<<"No\n";
			return 0;
		}
	}
	
	cout<<"Yes\n";
}