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>
#define REP(i, a, n) for(ll i = ((ll) a); i < ((ll) n); i++)
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;

ll N, M;
vector< vector<pll> > E;
vector<bool> used;
vector<ll> X;

bool dfs(ll v, ll x) {
  if(used[v]) return X[v] == x;
  used[v] = true;
  X[v] = x;

  bool ret = true;
  for(pll p : E[v]) {
    ll u = p.first, d = p.second;
    ret = ret && dfs(u, x + d);
  }
  return ret;
}

int main(void) {
  cin >> N >> M;
  E.resize(N);
  REP(i, 0, M) {
    ll L, R, D;
    cin >> L >> R >> D;
    L--;
    R--;
    E[L].push_back(pll(R, +D));
    E[R].push_back(pll(L, -D));
  }

  bool ans = true;
  used.resize(N, false);
  X.resize(N);
  REP(i, 0, N) if(!used[i]) ans = ans && dfs(i, 0);
  cout << (ans ? "Yes" : "No") << endl;
}