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 fi first
#define se second
using namespace std;
typedef long long ll;
const int N=1e5+5;
int n,q,f[N];
ll m;
int getf(int x){return x==f[x]?x:f[x]=getf(f[x]);}
vector<pair<int,int> >vc;
int main()
{
    scanf("%d%lld%d",&n,&m,&q);
    for(int i=1;i<=n;i++) f[i]=i;
    for(int i=1;i<=q;i++)
    {
        int u,v,c;scanf("%d%d%d",&u,&v,&c);
        u++;v++;
        if(!c) f[getf(u)]=getf(v);
        else vc.push_back({u,v});
    }
    for(int i=0;i<vc.size();i++)
        if(getf(vc[i].fi)==getf(vc[i].se)){printf("No\n");return 0;}
    int tot=0;
    for(int i=1;i<=n;i++) if(getf(i)==i) tot++;
    m-=n-tot;
    if((tot<=2||m==n-1)&&vc.size()) {printf("No\n");return 0;}
    if(m<=(ll)tot*(tot-1)/2) printf("Yes\n");
    else printf("No\n");
}