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 sz(v) ((int)(v).size())
using namespace std;
using lint = long long;
using pi = pair<int, int>;
const int MAXN = 100005;

struct disj{
	int pa[MAXN];
	void init(int n){
		iota(pa, pa + n + 1, 0);
	}
	int find(int x){
		return pa[x] = (pa[x] == x ? x : find(pa[x]));
	}
	bool uni(int p, int q){
		p = find(p);
		q = find(q);
		if(p == q) return 0;
		pa[q] = p; return 1;
	}
}disj;

int n; lint m; int q;
vector<pi>v;
int main(){
	scanf("%d %lld %d",&n,&m, &q);
	disj.init(n);
	int compCnt = n;
	while(q--){
		int a, b, c;
		scanf("%d %d %d",&a,&b,&c);
		if(c == 0){
			compCnt -= disj.uni(a, b);
		}
		else{
			v.emplace_back(a, b);
		}
	}
	for(auto &i : v){
		if(disj.find(i.first) == disj.find(i.second)){
			puts("No");
			return 0;
		}
	}
	if(m == n - 1){
		if(sz(v)) puts("No");
		else puts("Yes");
		return 0;
	}
	if(compCnt == 1){
		puts("No");
		return 0;
	}
	if(compCnt == 2){
		puts("No");
		return 0;
	}
	lint st = n - 1;
	lint ed = n - compCnt + 1ll * (compCnt) * (compCnt - 1) / 2;
	if(st <= m && m <= ed){
		puts("Yes");
	}
	else puts("No");
}