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 <cstring>
#define rep(a,b,c) for(int a=b;a<=c;a++)
const int maxn=1e5+1;
#define int long long
inline int max(int a,int b) { return a>b?a:b; }
inline int min(int a,int b) { return a<b?a:b; }
int n,m,q,mine,maxe,a[maxn],b[maxn],c[maxn],fa[maxn],fx,fy,ltk,num;
inline int find(int x) { return (fa[x]==x)?x:fa[x]=find(fa[x]); }
signed main() {
	scanf("%lld %lld %lld",&n,&m,&q);
	rep(i,1,n) fa[i]=i; ltk=n;
	rep(i,1,q) {
		scanf("%lld %lld %lld",&a[i],&b[i],&c[i]);
		a[i]++; b[i]++;
		if(!c[i]) {
			num++; fx=find(a[i]); fy=find(b[i]);
			if(fx==fy) continue;
			fa[fy]=fx; ltk--; mine++; maxe++;
		}
	}
	if(m==(n-1)) {
		if(num==q) puts("Yes");
		else puts("No");
		return 0;
	}
	if(m<(n-1)) { puts("No"); return 0; }
	rep(i,1,q) {
		if(c[i]) {
			fx=find(a[i]); fy=find(b[i]);
			if(fx==fy) { puts("No"); return 0; }
		}
	}
	mine+=ltk; maxe+=(((ltk-1)*ltk)>>1);
	if(mine<=m&&maxe>=m) puts("Yes");
	else puts("No");
}