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>
#define Hina ios_base::sync_with_stdio(false), cin.tie(0)
#define MAXN 100000
#define pb push_back
#define int long long
using namespace std;
 
int n, m, q, k;
int a, b, c;
int dsu[MAXN+5];
bool tf = 1, p;
vector < pair<int,int> > arr;
 
int finding(int a) {
    return a == dsu[a] ? a : dsu[a] = finding(dsu[a]);
    if (dsu[a] != a) dsu[a] = finding(dsu[a]);
    return dsu[a];
}
 
signed main() {
    Hina;
    cin >> n >> m >> q;
    for (int i = 0; i < n; i++) dsu[i] = i;
    k = n;
    for (int i = 0; i < q; i++) {
        cin >> a >> b >> c;
        if (finding(a) != finding(b) && c == 0) {
            dsu[finding(a)] = finding(dsu[b]);
            k--;
        } else if (c) {
            arr.pb({a, b});
        }
        if (c) p = 1;
    }
    for (int i = 0; i < arr.size(); i++) {
        if (finding(arr[i].first) == finding(arr[i].second)) {
            tf = 0;
        }
    }
    if (m > n-k + k*(k-1)/2/* || (k<=2 && p)*/ || (m==n-1 && p)) tf = 0;
    if (tf) cout << "Yes" << endl;
    else cout << "No" << endl;
}