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;
#define rep(i,n) for (int i=0;i<(n);i++)
#define Rep(i,a,b) for (int i=(a);i<=(b);i++)
#define rrep(i,n) for (int i=(n)-1;i>=0;i--)
#define rRep(i,a,b) for (int i=(a);i>=b;i--)
const int Maxn=100005;
int n,m;
vector <pair <int,int> > v[Maxn];
bool used[Maxn];
int a[Maxn];
bool dfs(int cur,int col)
{
	if (used[cur])
	{
		return a[cur]==col;
	}
	used[cur]=true;
	a[cur]=col;
	for (auto &u:v[cur])
	{
		if (!dfs(u.first,a[cur]+u.second))
		{
			return false;
		}
	}
	return true;
}
int main()
{
	cin>>n>>m;
	for (int i=0;i<m;i++)
	{
		int a,b,c;
		cin>>a>>b>>c;
		v[a].push_back(make_pair(b,c));
		v[b].push_back(make_pair(a,-c));
	}
	for (int i=1;i<=n;i++)
	{
		if (!used[i])
		{
			if (!dfs(i,0))
			{
				cout<<"No";
				return 0;
			}
		}
	}
	cout<<"Yes";
	return false;
}