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 <cstdio>
#include <algorithm>

const int MAXN = 100004;

int N;
int uf[MAXN];
long long xx[MAXN];

int findp(int);

int main() {
  int N, q;
  scanf("%d%d", &N, &q);
  std::fill(uf, uf + N, -1);
  for (int i = 0; i < q; ++i) {
    int l, r, d;
    scanf("%d%d%d", &l, &r, &d);

    int ll = findp(l), rr = findp(r);
    if (ll == rr) {
      if (xx[r] - xx[l] != d) goto fail;
      continue;
    }

    uf[rr] = ll;
    xx[rr] = (d + (long long)xx[l]) - xx[r];
  }
  puts("Yes");
  return 0;

fail:
  puts("No");
  return 0;
}

int Q[MAXN];
int qr;

int findp(int t) {
  Q[0] = t;
  qr = 1;
  int pt = uf[t];
  while (pt != -1) {
    t = Q[qr++] = pt;
    pt = uf[t];
  }
  int sum = 0;
  for (int i = qr - 2; i >= 0; --i) {
    sum += xx[Q[i]];
    xx[Q[i]] = sum;
    uf[Q[i]] = t;
  }
  return t;
}