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:
//by Judge
#include<bits/stdc++.h>
#define P pair<int,int>
#define fp(i,a,b) for(register int i=(a),I=(b)+1;i<I;++i)
#define ll long long
using namespace std;
const int M=1e5+3;
#ifndef Judge
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
#endif
char buf[1<<21],*p1=buf,*p2=buf;
inline ll read(){ ll x=0,f=1; char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0'; return x*f;
} int n,cnt,q,f[M]; ll m; vector<P> e[2];
inline void ask(bool x){ if(x) printf("No\n"),exit(0); }
int find(int x){ return x^f[x]?f[x]=find(f[x]):x; }
inline ll Sum(int x){return 1ll*x*(x-1)/2;}
inline void merge(int x,int y){
	x=find(x),y=find(y); if(x^y) ++cnt,f[y]=x;
}
int main(){ n=read(),m=read(),q=read(); fp(i,1,n) f[i]=i; int x,y,c;
	fp(i,1,q) x=read()+1,y=read()+1,c=read(),e[c].push_back(P(x,y));
	for(auto x: e[0]) merge(x.first,x.second);
	for(auto x: e[1]) ask(find(x.first)==find(x.second));
	ll Mn=e[1].size()?n:n-1,Mx=cnt+Sum(n-cnt);
	return ask(m>Mx||m<Mn),!printf("Yes\n");
}