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 max_n=100005;
typedef pair<int,int> P;
struct DSU{
    int par[max_n];
    int sz;
    void init(int n){
        for(int i=1;i<=n;i++)par[i]=i;
        sz=n;
    }
    int findpar(int x){
        return x==par[x]?x:par[x]=findpar(par[x]);
    }
    void unite(int x,int y){
        x=findpar(x),y=findpar(y);
        if(x!=y)par[x]=y,sz--;
    }
    bool same(int x,int y){
        return findpar(x)==findpar(y);
    }
}D;
int n;
ll m;
int q;
vector<P> check;
int main(){
    scanf("%d%lld%d",&n,&m,&q);
    D.init(n);
    while(q--){
        int u,v,w;scanf("%d%d%d",&u,&v,&w);
        if(w==0)D.unite(u,v);
        else check.push_back(P(u,v));
    }
    if(m==n-1){
        if(check.size()==0)printf("Yes\n");
        else printf("No\n");
        return 0;
    }
    for(auto p:check){
        int u=p.first,v=p.second;
        if(D.same(u,v)){
            printf("No\n");
            return 0;
        }
    }
    int k=D.sz;
    ll mx=n-k+1ll*k*(k-1)/2;
    if(m<=mx)printf("Yes\n");
    else printf("No\n");
    return 0;
}