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 N=1e5+5;
int fa[N];
int dis[N];
int n,m,x,y,d;
void init()
{
    for(int i=1;i<=n;i++)
        fa[i]=i,dis[i]=0;
}
int find(int x)
{
    if(fa[x]==x)return x;
    int k=fa[x];
    fa[x]=find(fa[x]);
    dis[x]+=dis[k];
    return fa[x];
}
void merge(int x,int y)
{
	int xx=find(x),yy=find(y);
    fa[yy]=xx;
    dis[yy]=dis[x]+d-dis[y];
}
int main()
{
    cin>>n>>m;
    init();
    while(m--)
    {
        cin>>x>>y>>d;
        if(find(x)!=find(y))
        {
            merge(x,y);
        }
        else{
            if(dis[y]-dis[x]!=d)
        	{
        		puts("No");
        		return 0;
        	}
        }
    }
    cout<<"Yes"<<endl;
    return 0;
}