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 <cstdio>
#include <iostream>
#define debug(...) fprintf(stderr, __VA_ARGS__)
using namespace std;
inline char nc() {
	return getchar();
	static char buf[100000], *l = buf, *r = buf;
	return l==r&&(r=(l=buf)+fread(buf,1,100000,stdin),l==r)?EOF:*l++;
}
template<class T> void read(T &x) {
	x = 0; int f = 1, ch = nc();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=nc();}
	while(ch>='0'&&ch<='9'){x=x*10-'0'+ch;ch=nc();}
	x *= f;
}
typedef long long ll;
const int maxn = 1e5 + 50;
ll m;
int n, Q;
int A[maxn], B[maxn];
bool C[maxn];
int k;
namespace us {
	int fa[maxn];
	void init() {
		for(int i = 1; i <= n; ++i) {
			fa[i] = i;
		}
	}
	int find(int a) {
		return a == fa[a] ? a : fa[a] = find(fa[a]);
	}
	bool merge(int a, int b) {
		a = find(a);
		b = find(b);
		if(a == b) return 0;
		fa[a] = b; return 1;
	}
}
bool solve() {
	if(m == n - 1) {
		for(int i = 1; i <= Q; ++i) {
			if(C[i] == 1) return 0;
		}
		return 1;
	}
	for(int i = 1; i <= Q; ++i) if(C[i] == 1) {
		if(us :: find(A[i]) == us :: find(B[i])) {
			return 0;
		}
	}
	return m <= n + (ll)k * (k - 3) / 2;
}
int main() {
	read(n), read(m), read(Q);
	k = n;
	us :: init();
	for(int i = 1; i <= Q; ++i) {
		read(A[i]), read(B[i]), read(C[i]);
		if(C[i] == 0) {
			if(us :: merge(A[i], B[i])) --k;
		}
	}
	puts(solve() ? "Yes" : "No");
	return 0;
}