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;
vector<pair<int,int>>adj[100005];
int dist[100005];
bool vis[100005];
bool dfs(int u, int cur){
  if(vis[u]){
    if(dist[u] != cur)
      return 0;
    return 1;
  }

  vis[u] = 1;
  dist[u] = cur;
  bool ret = 1;
  for(auto v : adj[u])ret &= dfs(v.first, cur + v.second);
  return ret;
}
int main(){
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M;

  while(M--){

    int u, v, w;
    cin >> u >> v >> w;

    adj[u].push_back(make_pair(v,w));
    adj[v].push_back(make_pair(u,-w));

  }
  bool f = 1;
  for(int i = 1; i <= N; i++)if(!vis[i])f &= dfs(i,0);
  cout  << ( f ? "Yes\n" : "No\n");
}