Problem: Snuke's mother gave Snuke an undirected graph consisting of N vertices numbered 0 to N-1 and M edges. This graph was connected and contained no parallel edges or self-loops.

One day, Snuke broke this graph. Fortunately, he remembered Q clues about the graph. The i-th clue (0 \leq i \leq Q-1) is represented as integers A_i,B_i,C_i and means the following:

* If C_i=0: there was exactly one simple path (a path that never visits the same vertex twice) from Vertex A_i to B_i.
* If C_i=1: there were two or more simple paths from Vertex A_i to B_i.



Snuke is not sure if his memory is correct, and worried whether there is a graph that matches these Q clues. Determine if there exists a graph that matches Snuke's memory.

Constraints

* 2 \leq N \leq 10^5
* N-1 \leq M \leq N \times (N-1)/2
* 1 \leq Q \leq 10^5
* 0 \leq A_i,B_i \leq N-1
* A_i \neq B_i
* 0 \leq C_i \leq 1
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N M Q
A_0 B_0 C_0
A_1 B_1 C_1
\vdots
A_{Q-1} B_{Q-1} C_{Q-1}


Output

If there exists a graph that matches Snuke's memory, print `Yes`; otherwise, print `No`.

Examples

Input

5 5 3
0 1 0
1 2 1
2 3 0


Output

Yes


Input

4 4 3
0 1 0
1 2 1
2 3 0


Output

No


Input

10 9 9
7 6 0
4 5 1
9 7 0
2 9 0
2 3 0
4 1 0
8 0 0
9 1 0
3 0 0


Output

No

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

#define rep(i, a, b) for (int i = (a), i##end = (b); i <= i##end; ++i)
#define per(i, a, b) for (int i = (a), i##end = (b); i >= i##end; --i)

#define int long long

const int maxn = 2e5 + 10;
int n, m, q, tot;

struct Querys {
  int op, u, v;
} Q[maxn];

int par[maxn];

int find(int x) {
  return par[x] == x ? x : par[x] = find(par[x]);
}

void unite(int x, int y) {
  if ((x = find(x)) != ((y = find(y)))) {
    par[x] = y, tot--;
  }
}

signed main() {
  scanf("%lld %lld %lld", &n, &m, &q);
  tot = n;
  rep(i, 1, n) par[i] = i;
  rep(i, 1, q) {
    int u, v, op;
    scanf("%lld %lld %lld", &u, &v, &op);
    Q[i] = Querys{op, ++u, ++v};
    if (!op) unite(u, v);
    if (m == n - 1 && Q[i].op) {
      return puts("No"), 0;
    }
  }
  if (m == n - 1) {
    return puts("Yes"), 0;
  }
  rep(i, 1, q) {
    if (Q[i].op && find(Q[i].u) == find(Q[i].v)) {
      return puts("No"), 0;
    }
  }
  if (m >= n && m <= n - tot + tot * (tot - 1) / 2) {
    puts("Yes");
  } else {
    puts("No");
  }
  return 0;
}