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 ll long long
#define PII pair < int , int >
const int _ = 1e5 + 7;
ll fa[_] , sz[_] , N , M , Q;
vector < PII > to0 , to1;
int find(int x){return fa[x] == x ? x : (fa[x] = find(fa[x]));}
void merge(int p , int q){if((p = find(p)) == (q = find(q))) return; fa[p] = q; sz[q] += sz[p];}
signed main(){
	ios::sync_with_stdio(0);
	cin >> N >> M >> Q;
	for(int i = 1 ; i <= N ; ++i) sz[fa[i] = i] = 1;
	for(int i = 1 ; i <= Q ; ++i){
		int A , B , C; cin >> A >> B >> C; ++A; ++B;
		if(C == 0) to0.push_back(PII(A , B));
		else to1.push_back(PII(A , B));
	}
	for(auto t : to0) merge(t.first , t.second);
	for(auto t : to1) if(find(t.first) == find(t.second)){puts("No"); return 0;}
	ll mn = 0 , mx = 0 , cnt = 0; bool flg = 1;
	for(int i = 1 ; i <= N ; ++i) if(find(i) == i){++cnt; mn += sz[i] - 1; mx += sz[i] - 1;}
	for(int i = 1 ; i <= N ; ++i) if(fa[i] == i) sz[i] = 1; else sz[i] = 0;
	for(auto t : to1) merge(t.first , t.second);
	for(int i = 1 ; i <= N ; ++i)
		if(fa[i] == i && sz[i] == 2 && cnt == 2){puts("No"); return 0;}
		else if(sz[i] >= 2) flg = 0;
	mn += cnt - flg; if(M < mn){puts("No"); return 0;}
	mx += cnt * (cnt - 1) / 2; if(M > mx){puts("No"); return 0;}
	puts("Yes"); return 0;
}