#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e5+5;
int l[N], r[N], op[N],fa[N];
inline int find(int x){return fa[x] == x ? fa[x] : fa[x] = find(fa[x]);}
int n,q;ll m;
int main()
{
	cin >> n >> m;
	for(int i=1;i<=n;i++)fa[i] = i;
	ll rest = n;
	cin >> q;
	for(int i=1;i<=q;i++){
		scanf("%d%d%d",&l[i], &r[i], &op[i]);
		++l[i], ++r[i];
		if(op[i]==0){
			int x = find(l[i]), y = find(r[i]);
			if(x^y){
				fa[x] = y;
				rest --;
			}
		}
	}
	for(int i=1;i<=q;i++){
		if(op[i]==1){
			int x = find(l[i]), y = find(r[i]);
			if(x==y || m==n-1){
				puts("No");
				return 0;
			}
		}
	}
	ll Max = rest*(rest-1)/2 + n-rest;
	if(Max >= m){
		puts("Yes");
	}else puts("No");
}
### Prompt

Your task is to create a Cpp solution to the following 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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e5+5;
int l[N], r[N], op[N],fa[N];
inline int find(int x){return fa[x] == x ? fa[x] : fa[x] = find(fa[x]);}
int n,q;ll m;
int main()
{
	cin >> n >> m;
	for(int i=1;i<=n;i++)fa[i] = i;
	ll rest = n;
	cin >> q;
	for(int i=1;i<=q;i++){
		scanf("%d%d%d",&l[i], &r[i], &op[i]);
		++l[i], ++r[i];
		if(op[i]==0){
			int x = find(l[i]), y = find(r[i]);
			if(x^y){
				fa[x] = y;
				rest --;
			}
		}
	}
	for(int i=1;i<=q;i++){
		if(op[i]==1){
			int x = find(l[i]), y = find(r[i]);
			if(x==y || m==n-1){
				puts("No");
				return 0;
			}
		}
	}
	ll Max = rest*(rest-1)/2 + n-rest;
	if(Max >= m){
		puts("Yes");
	}else puts("No");
}
```