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, n) for (int i = 0; i < (int)(n); i++)
#define rep_in(i, s, n) for (int i = (s); i < (int)(n); i++)
typedef int64_t ll;


vector<int> si;
int root(int i){
  if(si[i]==i) return i;
  return si[i]=root(si[i]);
}

bool solve() {
  int n;long m;int q;
  cin>>n>>m>>q;
  vector<int> a(q),b(q),c(q);
  rep(i,q){
    cin>>a[i]>>b[i]>>c[i];
  }

  rep(i,n){
    si.push_back(i);
  }

  int islands=n;
  rep(i,q){
    if(c[i]!=0) continue;
    int ra=root(a[i]);
    int rb=root(b[i]);
    if(ra!=rb) {
      si[ra]=rb;
      islands--;
    }
  }

  int doubles=0;
  rep(i,q){
    if(c[i]!=1) continue;
    doubles++;
    int ra=root(a[i]);
    int rb=root(b[i]);
    if(ra==rb) return false;
  }

  if(doubles>0&&m==n-1) return false; // 輪を作るのにエッジが足りない
  //if(doubles>0&&islands<3) return false; // 輪を作るのにノードが足りない
  if(m>(n-1)+((long)islands*(islands-1)/2-(islands-1))) return false;
  return true;
}

int main() {
  cout << (solve()?"Yes":"No") << endl;
}