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;
const int maxn=2e5+10;
struct edge{int u,v,typ;}a[maxn+3];
int fa[maxn+3];
int n,q,c; 
long long m;
int find(int u){return fa[u]==u?u:fa[u]=find(fa[u]);}
int main()
{
    scanf("%d%lld%d",&n,&m,&q);
    for(int i=1;i<=n;i++)fa[i]=i;
    for(int i=1;i<=q;i++)
    {
        scanf("%d%d%d",&a[i].u,&a[i].v,&a[i].typ);
		a[i].u++,a[i].v++;
        if(a[i].typ==0)
        {
            int u=find(a[i].u),v=find(a[i].v);
            if(u!=v)fa[u]=v;
        }
    }
    int flag=0;c=0;
    for(int i=1;i<=n;i++)if(i==find(i))c++;
    for(int i=1;i<=q;i++)
    if(a[i].typ==1)
    {
        flag=1;
        int u=find(a[i].u),v=find(a[i].v);
        if(u==v){puts("No");return 0;}
    }
    long long l=flag?max(3,c):c-1,r=c*1LL*(c-1ll)/2ll;
    if(m-(n-c)>=l&&m-(n-c)<=r)puts("Yes");
    else puts("No");
    return 0;
}