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 N = 100010;

vector < pair <int, long long> > g[N];
bool used[N];
long long d[N];

long long mx;

void dfs(int x) {
  used[x] = 1;
  for (auto &e : g[x]) {
    int to = e.first;
    int dist = e.second;
    if (used[to] && d[to] != d[x] + dist) {
      puts("No");
      exit(0);
    } else if (!used[to]) {
      d[to] = d[x] + dist;
      mx = max(mx, d[to]);
      dfs(to);
    }
  }
}

int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  int st = 0;
  for (int i = 0; i < m; ++i) {
    int x, y, z;
    scanf("%d %d %d", &x, &y, &z);
    g[x].push_back({y, z});
    g[y].push_back({x, -z});
    st = x;
  }
  mx = -1;
  for (int i = 1; i <= n; ++i) {
    if (!used[i]) {
      ++mx;
      dfs(i);
    }
  }
  puts("Yes");
	return 0;
}