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 <iostream>
using namespace std;

const int N = 1e5 + 5;

int fa[N], val[N];

inline int getf(int x) {
  if (x == fa[x]) return x;
  else {
    getf(fa[x]);
    val[x] += val[fa[x]];
    fa[x] = fa[fa[x]];
    return fa[x];
  }
}

int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i ++) fa[i] = i;
  for (int a, b, l, r, d; m; m --) {
    scanf("%d%d%d", &l, &r, &d);
    a = getf(l); b = getf(r);
    if (a == b) {
      if (val[r] - val[l] != d) return puts("No"), 0;
    }
    else {
      d += val[l] - val[r];
      fa[b] = a;
      val[b] += d;
    }
  }
  puts("Yes");
  return 0;
}