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;
#define rep(i,s,t) for(int i=s;i<t;i++)
#define pii pair<int,int>
typedef long long ll;
#define MAXNUM 111111
vector<pii> v[2];
int dset[MAXNUM],ssize[MAXNUM];
int fa(int a)
{
    if(dset[a]!=a)return dset[a]=fa(dset[a]);
    return a;
}
int main()
{
    ll m;
    int n,q,a,b,c;scanf("%d%lld%d",&n,&m,&q);
    rep(i,1,n+1)dset[i]=i;
    rep(i,1,q+1)scanf("%d%d%d",&a,&b,&c),a++,b++,v[c].emplace_back(a,b);
    ll up=n*1ll*(n-1)/2,down=n-1;
    int flag=1;
    for(pii p:v[0])
        dset[fa(p.first)]=fa(p.second);
    int dnum=0;
    rep(i,1,n+1)ssize[fa(i)]++,dnum+=i==fa(i);
    for(pii p:v[1])
    {
        if(fa(p.first)==fa(p.second))flag=0;
    }
    if(!v[1].empty())down++;
    if(!v[1].empty()&&dnum<=2)flag=0;
    up=dnum*1ll*(dnum-1)/2;
    rep(i,1,n+1)if(fa(i)==i)
        up+=ssize[i]-1;
    if(m>up||m<down)flag=0;
    if(flag)printf("Yes\n");
    else printf("No\n");
}