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>
#pragma GCC optimize("O3")
using namespace std;
#ifdef _debug
#define dout(i) cout << #i << ' ' << i << ' '
#else 
#define dout(i) //
#endif
using ll = long long;
using ull = unsigned long long;
using ul = unsigned;
using db = double;
int n, q;
ll m, c;
vector<pair<int,int>> edge[2];
const int maxn = 100001;
int s[maxn], b[maxn];
string res[2] = {"No\n", "Yes\n"};
int find(int i){
	return i == b[i] ? i : b[i] = find(b[i]);
}
void merge(int x, int y){
	x = find(x), y = find(y);
	if(s[x] < s[y])swap(x, y);
	if(x != y){
		s[x] += s[y];
		b[y] = b[x];
		--c;
	}
}
#define same(a, b) (find(a) == find(b))
signed main(){
	ios_base::sync_with_stdio(0), cin.tie(0);
	cin >> n >> m >> q;
	for(int i = 0, a, b, c;i < q;++i){
		cin >> a >> b >> c;
		edge[c].emplace_back(a, b);
	}
	if(m == n-1)return cout << res[edge[1].empty()], 0;
	c = n;
	for(int i = 1;i <= n;++i)b[i] = i, s[i] = 1;
	for(auto &p : edge[0])merge(p.first, p.second);
	for(auto &p : edge[1])if(same(p.first, p.second))return cout << res[0] , 0;
	cout << res[m <= n-c + c*(c-1)/2];
}