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 ll long long
using namespace std;
const int N = 2e5;
struct Element
{
    int u,v,opt;
} p[N+3];
int fa[N+3];
int n,q,c; ll m;

int find(int u) {return fa[u]==u?u:fa[u]=find(fa[u]);}

int main()
{
    scanf("%d%lld%d",&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 uu = find(p[i].u),vv = find(p[i].v);
            if(uu!=vv) {fa[uu] = vv;}
        }
    }
    int huan = 0; c = 0;
    for(int i=1; i<=n; i++) {if(i==find(i)) c++;}
    for(int i=1; i<=q; i++){
        if(p[i].opt==1){
            huan++;
            int uu = find(p[i].u),vv = find(p[i].v);
            if(uu==vv) {puts("No"); return 0;}
        }
    }
    ll l = huan?max(3ll,(ll)c):c-1ll,r = 1ll*c*(c-1ll)/2ll;
    if(m-(n-c)>=l && m-(n-c)<=r) {puts("Yes");}
    else {puts("No");}
    return 0;
}