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 = 100010, INF = (int) 1e9 + 10;
vector< pair<int,int> > edge[N];
int d[N];

bool DFS(int v, int dis) {
  if ( d[v] != -INF && d[v] != dis ) return 0;
  else if ( d[v] != -INF && d[v] == dis ) return 1;
  d[v] = dis;
  bool ok = 1;
  for ( auto u:edge[v] ) {
    ok &= DFS(u.first,dis+u.second);
  }
  return ok;
}

int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; i++) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    edge[u].push_back( make_pair(v,w) );
    edge[v].push_back( make_pair(u,-w) );
  }
  fill( d+1, d+n+1, -INF);
  bool ok = 1;
  for (int i = 1; i <= n; i++) {
    if ( d[i] == -INF ) {
      ok &= DFS(i,0);
    }
  }
  puts( ok ? "Yes" : "No");
  return 0;
}