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;
typedef long long ll;
const int maxn=2e5+5;
struct Node{
    int u,v,opt;
}p[maxn];
int fa[maxn];
int find(int x){
    return x==fa[x]?x:fa[x]=find(fa[x]);
}
int main(){
    ll n,m,q;
    cin>>n>>m>>q;
    for(int i=1;i<=n;i++) fa[i]=i;
    for(int i=1;i<=q;i++){
        scanf("%d%d%d",&p[i].u,&p[i].v,&p[i].opt);
        p[i].u++;
        p[i].v++;
        if(p[i].opt==0){
            int fu=find(p[i].u),fv=find(p[i].v);
            if(fu!=fv)
                fa[fu]=fv;
        }
    }
    ll c=0;
    for(int i=1;i<=n;i++){
        if(find(i)==i)
            c++;
    }
    ll huan=0;
    for(int i=1;i<=q;i++){
        if(p[i].opt==1){
            huan++;
            if(find(p[i].u)==find(p[i].v)){
                printf("No\n");
                return 0;
            }
        }
    }
    ll temp=m-(n-c);
    ll l,r;
    l = huan?max(3ll,(ll)c):c-1ll,r = 1ll*c*(c-1ll)/2ll;
    if(l<=temp&&temp<=r)
        printf("Yes\n");
    else printf("No\n");
}