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;
typedef long long ll;
typedef pair<int, int> pii;
#define F first
#define S second
const int maxn = 2e5 + 10;
int n, q;
ll m;
int p[maxn], sz[maxn];
int find(int x) {return p[x] == x ? x : p[x] = find(p[x]);}
vector<pii> edges[2];
bool Union(int x, int y) {
	x = find(x), y = find(y);
	if(x == y) return 0;
	if(sz[x] < sz[y]) swap(x, y);
	p[y] = x, sz[x] += sz[y];
	return 1;
}
int main() {
	scanf("%d%lld%d", &n, &m, &q);
	for(int i = 0; i < q; i++) {
		int x, y, c;
		scanf("%d%d%d", &x, &y, &c);
		edges[c].push_back({x, y});
	}
	if(m == n - 1 && !edges[1].empty())
		return puts("No"), 0;
	
	for(int i = 0; i < n; i++) p[i] = i, sz[i] = 1;
	int com = n;
	for(auto p : edges[0]) 
		if(Union(p.F, p.S)) com--;
	for(auto p : edges[1]) 
		if(find(p.F) == find(p.S)) return puts("No"), 0;

	if(com != 2 && m <= n - com + 1LL * com * (com - 1) / 2) 
		puts("Yes");
	else puts("No");
	return 0;
}