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 pii pair <int , int>
#define F first
#define S second

const int N = 1e5 + 5;
int par[N];
long long C, m;

int getpar(int v) {return par[v] = (par[v] == v ? v : getpar(par[v]));}
void Union(int u, int v) {
    u = getpar(u), v = getpar(v);
    if (u == v) return ;
    C--, par[u] = par[v];
    return ;
}

int main(){
    int n, q; cin >> n >> m >> q, C = n;
    vector <pii> vec;
    for (int i = 0; i < n; i++) par[i] = i;
    for (int i = 0; i < q; i++) {
        int u, v, c; cin >> u >> v >> c;
        if (c == 0) Union(u, v);
        else vec.push_back({u, v});
    }
    for (pii p : vec) if (getpar(p.F) == getpar(p.S)) return cout << "No\n", 0;
    if ((m == n - 1 || C == 2) && !vec.empty()) return cout << "No\n", 0;
    if (m > C * (C - 1) / 2 + n - C) return cout << "No\n", 0;
    cout << "Yes\n";

    return 0;
}