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;
mt19937 rng(std::chrono::system_clock::now().time_since_epoch().count());
const int N = 1e5;
int n,q,p[N],ans;
long long m;
bool is;
vector<pair<int,int>>bad;
int get(int u){return u == p[u] ? u : p[u] = get(p[u]);}
int main(){
	//freopen("input.txt","r",stdin);
	scanf("%d%lld%d",&n,&m,&q);
	ans = n;
	for(int i = 0;i < n;i++)p[i] = i;
	for(int i = 0,a,b,c;i < q;i++){
		scanf("%d%d%d",&a,&b,&c);
		if(c == 1){
			is = 1;
			bad.push_back({a,b});
			continue;
		}
		if(get(a) != get(b))ans--;
		p[get(a)] = get(b);
	}
	for(auto i : bad)
		if(get(i.first) == get(i.second)){
			puts("No");
			return 0;
		}
	if((ans < 3 || m == n - 1) && is){
		puts("No");
		return 0;
	}
	puts(n + ans*(ans - 3LL)/2 >= m ? "Yes" : "No");
}