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;

int n,m,fa[202000],d[202000];

int GF(int x){
  if (fa[fa[x]]==fa[x]) return fa[x];
  int orgfa=fa[x];
  fa[x]=GF(fa[x]);
  d[x]+=d[orgfa];
  return fa[x];
}
int dis(int x){
  return fa[x]==x? 0: d[x]+dis(fa[x]);
}

void onion(int x,int y,int v){
  int fx=GF(x), dx=dis(x), fy=GF(y), dy=dis(y), las;
  las=-dx+dy+v;
  if (fx==fy){
    if (las){
      puts("No");
      exit(0);
    }
  }else{
    fa[fx]=fy;
    d[fx]=las;
  }
}

int main(){
  cin>>n>>m;
  for (int i=1;i<=n;++i)
    fa[i]=i, d[i]=0;
  int l,r,x;
  for (;m--;){
    scanf("%d%d%d",&l,&r,&x);
    onion(r,l,x);
  }
  puts("Yes");
}