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:
// D.cpp
#include <bits/stdc++.h>

using namespace std;

const int MAX_N = 1e5 + 200;

int n, m, q, mem[MAX_N], siz[MAX_N];

struct clue
{
	int a, b, c;
} clues[MAX_N];

int find(int x) { return mem[x] == x ? x : mem[x] = find(mem[x]); }

int main()
{
	scanf("%d%d%d", &n, &m, &q);
	for (int i = 1; i <= n; i++)
		mem[i] = i;
	for (int i = 1; i <= q; i++)
	{
		scanf("%d%d%d", &clues[i].a, &clues[i].b, &clues[i].c);
		clues[i].a++, clues[i].b++;
		if (clues[i].c == 0 && find(clues[i].a) != find(clues[i].b))
			mem[find(clues[i].a)] = find(clues[i].b);
	}
	int component = 0, c1 = 0;
	for (int i = 1; i <= n; i++)
		component += find(i) == i;
	for (int i = 1; i <= q; i++)
		if (clues[i].c == 1)
		{
			c1++;
			int u = find(clues[i].a), v = find(clues[i].b);
			if (u == v)
				puts("No"), exit(0);
		}
	long long lft = c1 ? max(3, component) : component - 1, rig = 1LL * component * (component - 1) / 2;
	puts((lft <= m - (n - component) && m - (n - component) <= rig) ? "Yes" : "No");
	return 0;
}