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 <stdio.h>
#include <string.h>
using namespace std;
const int N=1000005;
int n,m;
int fa[N],dis[N];
int finds(int x) {
  if(x==fa[x]) return x;
  int ret=fa[x];
  fa[x]=finds(fa[x]);
  dis[x]+=dis[ret];
  return fa[x];
}
int main() {
  scanf("%d%d",&n,&m);
  for (int i=1;i<=n;++i) fa[i]=i;
  for (int i=1;i<=m;++i) {
    int x,y,d;
    scanf("%d%d%d",&x,&y,&d);
    int fx=finds(x),fy=finds(y);
    if(fx==fy && (dis[x]-dis[y])!=d) {
      puts("No");
      return 0;
    }
    if(fx!=fy) {
      fa[fx]=fy;
      dis[fx]=-dis[x]+dis[y]+d;
    }
  }
  puts("Yes");
  return 0;
}