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 <cstdio>
#include <vector>
#include <utility>
using namespace std;
typedef pair<int,int> ii;

int n,q;
long long m;

int p[100005];
long long components;

int parent(int i){return (p[i]==i)?i:p[i]=parent(p[i]);}

vector<ii> diff;

int main(){
	for (int x=0;x<100005;x++) p[x]=x;
	
	scanf("%d%lld%d",&n,&m,&q);
	components=n;
	
	int a,b,c;
	while (q--){
		scanf("%d%d%d",&a,&b,&c);
		if (c==0){
			a=parent(a),b=parent(b);
			if (a!=b){
				components--;
				p[a]=b;
			}
		}
		else{
			diff.push_back(ii(a,b));
		}
	}
	
	if (!diff.empty() && components<=2){
		printf("No");
		return 0;
	}
	
	for (auto &it:diff){
		if (parent(it.first)==parent(it.second)){
			printf("No");
			return 0;
		}
	}
	
	long long lo=n-(diff.empty()),hi=(components*(components-1))/2+n-components;
	
	//printf("%d %d %d\n",lo,hi,components);
	
	printf((lo<=m && m<=hi)?"Yes":"No");
}