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 MN 100000
#define ll long long
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
ll m;
pair<int,int> a[MN+5];
int n,Q,s[MN+5],num,ans;
inline int getfa(int x){return s[x]?s[x]=getfa(s[x]):x;}
inline void Merge(int x,int y){if(getfa(x)!=getfa(y)) s[getfa(x)]=getfa(y);}
int main()
{
    n=read();cin>>m;Q=read();
    for(int i=1;i<=Q;++i)
    {
        int x=read()+1,y=read()+1,op=read();
        if(op==0) Merge(x,y);
        else a[++num]=make_pair(x,y);
    }
    for(int i=1;i<=num;++i) 
        if(getfa(a[i].first)==getfa(a[i].second)) return 0*puts("No");
    if(m==n-1) return 0*puts(num?"No":"Yes");
    for(int i=1;i<=n;++i) if(getfa(i)==i) ++ans;
    if(ans==2)
    {
        if(num) return 0*puts("No");
        else if(m>n-1) return 0*puts("No");
        else return 0*puts("Yes");
    }
    long long max=1LL*ans*(ans-1)/2;
    if(n-ans+max<m) return 0*puts("No");
    puts("Yes");
    return 0;
}