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 <iostream>
#include <algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<iomanip>
#define sqr(x) (x)*(x)
using namespace std;
long long n,m,i,j,dis[100005],x,y,z;
vector<pair<long long,long long>/**/> e[100005];
void dfs(long long x,long long dist)
{
	if (dis[x])
	{
		if (dis[x]!=dist)
		{
			cout<<"No";
			exit(0);
		}
		return;
	}
	dis[x]=dist;
	int i;
	for (i=0;i<e[x].size();i++)
	{
		dfs(e[x][i].first,dist+e[x][i].second);
	}
}
int main()
{
	cin>>n>>m;
	while (m--)
	{
		cin>>x>>y>>z;
		e[x].push_back(make_pair(y,z));
		e[y].push_back(make_pair(x,-z));
	}
	for (i=1;i<=n;i++)
	{
		if (dis[i]) continue;
		dfs(i,1e15);
	}
	cout<<"Yes";
	return 0;
}