#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")

#include "bits/stdc++.h"
using namespace std;

#define pb push_back
#define F first
#define S second
#define f(i,a,b)  for(int i = a; i < b; i++)

using ll = long long;
using db = long double;
using ii = pair<int, int>;

const int N = 2e5 + 5, LG = 19, MOD = 1e9 + 7;
const int SQ =225;
const long double EPS = 1e-7;
vector<int>p;
int get(int i){return i == p[i] ? i : p[i]=get(p[i]);}
int cnt =0;
void mrg(int u, int v){
  if(get(u)==get(v))return;
  cnt++;
  p[get(v)] = get(u);
}
int32_t main(){
#ifdef ONLINE_JUDGE
    ios_base::sync_with_stdio(0);
    cin.tie(0);
#endif

  int n, q;
  ll m;

  cin >> n >> m >> q;
  p.resize(n);
  iota(p.begin(),p.end(),0);
  vector<ii> nonBridges;

  f(i,0,q){
    int u, v, c;
    cin >> u >> v >> c;
    if(c)
      nonBridges.pb(ii(u,v));
    else
      mrg(u,v);
  }

  if(m == n - 1){
    if(nonBridges.size())
      cout << "No\n";
    else
      cout << "Yes\n";
    return 0;
  }

  for(auto x : nonBridges)
  if(get(x.F)==get(x.S)){
    cout << "No\n";
    return 0;
  }

  if(m <= cnt + 1ll*(n-cnt)*(n-cnt-1)/2)
    cout<<"Yes\n";
  else
    cout<<"No\n";


  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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

### Response

```cpp
#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")

#include "bits/stdc++.h"
using namespace std;

#define pb push_back
#define F first
#define S second
#define f(i,a,b)  for(int i = a; i < b; i++)

using ll = long long;
using db = long double;
using ii = pair<int, int>;

const int N = 2e5 + 5, LG = 19, MOD = 1e9 + 7;
const int SQ =225;
const long double EPS = 1e-7;
vector<int>p;
int get(int i){return i == p[i] ? i : p[i]=get(p[i]);}
int cnt =0;
void mrg(int u, int v){
  if(get(u)==get(v))return;
  cnt++;
  p[get(v)] = get(u);
}
int32_t main(){
#ifdef ONLINE_JUDGE
    ios_base::sync_with_stdio(0);
    cin.tie(0);
#endif

  int n, q;
  ll m;

  cin >> n >> m >> q;
  p.resize(n);
  iota(p.begin(),p.end(),0);
  vector<ii> nonBridges;

  f(i,0,q){
    int u, v, c;
    cin >> u >> v >> c;
    if(c)
      nonBridges.pb(ii(u,v));
    else
      mrg(u,v);
  }

  if(m == n - 1){
    if(nonBridges.size())
      cout << "No\n";
    else
      cout << "Yes\n";
    return 0;
  }

  for(auto x : nonBridges)
  if(get(x.F)==get(x.S)){
    cout << "No\n";
    return 0;
  }

  if(m <= cnt + 1ll*(n-cnt)*(n-cnt-1)/2)
    cout<<"Yes\n";
  else
    cout<<"No\n";


  return 0;
}

```