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
#define maxn 1000005 /*rem*/
#define mod 998244353
#define db double
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define pi pair<int, int>
#define fi first
#define se second
using namespace std;
ll ksm(ll a, ll b) {
   if (!b) return 1;
   ll ns = ksm(a, b >> 1);
   ns = ns * ns % mod;
   if (b & 1) ns = ns * a % mod;
   return ns;
}
int fa[maxn];
int gfa(int a) {
	if (fa[a] == a) return a;
	return fa[a] = gfa(fa[a]);
}
int n;
ll m;
int q;
int a[maxn], b[maxn], c[maxn];
int main() {
	cin >> n >> m >> q;
	for (int i = 0; i < n; i++) fa[i] = i;
	for (int i = 0; i < q; i++) {
		scanf("%d%d%d", &a[i], &b[i], &c[i]);
		if (c[i] == 0) fa[gfa(a[i])] = gfa(b[i]);
	}
	int fl = 0, afl = 1;
	for (int i = 0; i < q; i++) {
		if (c[i] == 0) continue;
		if (gfa(a[i]) == gfa(b[i])) afl = 0;
		fl = 1;
	}
	int tot = 0;
	for (int i = 0; i < n; i++)
		if (fa[i] == i) tot++;
	ll l = n, r = n - tot + 1ll * tot * (tot - 1) / 2;
	if (!fl) l--;
	if (m > r || m < l) afl = 0;
	if (afl) cout << "Yes" << endl;
	else cout << "No" << endl;
	return 0;
}