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<iostream>
#include<cstdio>
#include<vector>
typedef std::pair<int,int> P;
const int N=1e5+5;
int par[N],n,q,cnt;
long long m;
void init() {
    for(int i=0;i<n;i++) par[i]=-1;
    cnt=n;
}
int fd(int x) {
    return par[x]==-1?x:(par[x]=fd(par[x]));
}
void unite(int x,int y) {
    x=fd(x);y=fd(y);
    if(x==y) return;
    --cnt;
    par[x]=y;
}
int main() {
    scanf("%d%lld%d",&n,&m,&q);
    init();
    std::vector<P> g;
    for(int u,v,c;q--;) {
        scanf("%d%d%d",&u,&v,&c);
        if(c==0) unite(u,v);
        else g.push_back(P(u,v));
    }
    if(m==n-1) {
        printf("%s",g.size()>0?"No":"Yes");
        return 0;
    }
    for(auto &c:g) if(fd(c.first)==fd(c.second)) {
        printf("No");
        return 0;
    }
    printf("%s",m<=n-cnt+1LL*cnt*(cnt-1)/2?"Yes":"No");
    return 0;
}