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<cstdio>
#include<algorithm>
using namespace std;
int n,m,cnt,tot;
int fns[100010],wz[100010],b[100010],roadl[400010],roadto[400010],road[400010];


void dfs(int num,int st)
{
  if(b[num])
    {
      if(st!=fns[num]) cnt=1;
      return;
    }
  b[num]=1,fns[num]=st;
  int t=wz[num];
  while(t)
    {
      dfs(road[t],st+roadl[t]);
      t=roadto[t];
    }
}

int main()
{ 
  int i,j,x,y,z;
  scanf("%d%d",&n,&m);
  for(i=1;i<=m;i++) 
  {
    scanf("%d%d%d",&x,&y,&z);
    roadto[++tot]=wz[x],road[tot]=y,wz[x]=tot,roadl[tot]=z;
    roadto[++tot]=wz[y],road[tot]=x,wz[y]=tot,roadl[tot]=-z;
  }
  for(i=1;i<=n;i++)
    {
      if(!b[i]) dfs(i,0);
    }
  if(cnt) printf("No\n");
  else printf("Yes\n");
}