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 MAXN = 100000, unknown = 1e9 + 100;

vector<pair<int,int>> g[MAXN];
int cur[MAXN];

bool dfs(int v) {
  bool ok = 1;
  for (auto P : g[v]) {
    int u = P.first, d = P.second;
    if (cur[u] == unknown) cur[u] = cur[v] + d, ok &= dfs(u);
    else {
      if (cur[u] != cur[v] + d) ok = 0;
    }
  }
  return ok;
}
int main() {
  int N, M, u, v, d; scanf("%d%d", &N, &M);
  for (int i = 0; i < M; ++i) scanf("%d%d%d", &u, &v, &d), --u, --v, g[u].push_back({v, d}), g[v].push_back({u, -d});
  fill(cur, cur + N, unknown);
  bool ok = 1;
  for (int i = 0; i < N; ++i) 
    if (cur[i] == unknown) 
      cur[i] = 0, ok &= dfs(i);
  puts(ok? "Yes": "No");
}