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 ll long long
using namespace std;
const int N = 2e5 + 10;
int f[N];
ll dis[N];

void fail() {
  puts("No");
  exit(0);
}

int find(int k) {
  if(f[k] == k) return k;
  else {
    int fa = find(f[k]);
    dis[k] += dis[f[k]];
    return f[k] = fa;
  }
}

int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for(int i = 1; i <= n; ++i) f[i] = i;
  for(int i = 1, x, y, z; i <= m; ++i) {
    scanf("%d%d%d", &x, &y, &z);
    int fx = find(x), fy = find(y);
    if(fx == fy) {
      if(dis[y] - dis[x] != z) fail();
    } else {
      f[fy] = fx;
      dis[fy] = dis[x] - dis[y] + z;
    }
  }
  puts("Yes");
  return 0;
}