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 <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const int INF=(int)(1<<30);
vector<pair<int, int> > graph[101010];
int vis[101010];
int d[101010];

void dfs(int cur, long long S) {
  if(vis[cur] == 1) {
    if(d[cur] != S) {
      cout << "No" << endl;
      exit(0);
    }
    return;
  }
  vis[cur] = 1;
  d[cur] = S;
  for(auto g:graph[cur]) dfs(g.first, S+g.second);
}

int main() {
  int N, M; cin >> N >> M;
  vector<int> L(M), R(M), D(M);
  for(int i = 0; i < M; ++i) {
    cin >> L[i] >> R[i] >> D[i];
    L[i]--, R[i]--;
    graph[L[i]].push_back(make_pair(R[i], D[i]));
    graph[R[i]].push_back(make_pair(L[i], -D[i]));
  }
  for(int i = 0; i < N; ++i) {
    if(vis[i] == 0) dfs(i, 0);
  }
  cout << "Yes" << endl;
  return 0;
}