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>
#define pb push_back
#define ins insert
#define F first
#define S second
#define var auto
using namespace std;
typedef long long ll;
const int Max = 1e5 + 10;
const int Mod = 1e9 + 7;


bool seen[Max];
vector<pair<int,int> > N[Max];

bool ok = true;
int dist[Max];

void DFS(int v)
{
	seen[v] = true;

	for(auto u : N[v])
	{
		if(seen[u.F])
			ok &= dist[u.F] == dist[v] + u.S;
		else
		{
			dist[u.F] = dist[v] + u.S;
			DFS(u.F);
		}
	}
}

int main()
{
	int n , m;cin >> n >> m;
	for(int i = 0; i < m ; i++)
	{
		int l , r , d;cin >> l >> r >> d;
		N[l].pb({r , d});
		N[r].pb({l , -d});
	}
	for(int i = 1; i <= n ; i++)
	{
		if(!seen[i])
		{
			dist[i] = 0;
			DFS(i);
		}
	}
	if(ok)
		cout << "Yes\n";
	else
		cout << "No\n";
}