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:
//#pragma GCC optimize("trapv")
#include<bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define pb push_back
using namespace std;
using ll = long long;
using vi = vector<ll>;
using pi = pair<ll, ll>;
using vpi = vector<pi>;
const int maxn = 3e5 + 33, mod = 1e9 + 7;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void quit() {cout << "No"; exit(0);}
struct dsu {
	vi r, p;
	int k;
	void init(int n) {
		k = n;
		r.resize(n);
		p.resize(n);
		for(int i = 0; i <n ; i++) p[i] = i;
	}
	int par(int i) {
		if(i==p[i]) return i;
		return p[i] = par(p[i]);
	}
	void unite(int i, int j) {
		i = par(i), j = par(j);
		if(i==j) return;
		if(r[i] < r[j]) swap(i, j);
		p[j] = i;
		r[i] += r[i] == r[j];
		--k;
	}
	bool con(int i, int j) {return par(i) == par(j);}
};
ll n, m, q;
vpi a[2];
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin >> n >> m >> q;
	for(int x, b, c, i = 0; i < q; i++) {
		cin >> x >> b >> c;
		a[c].pb({x, b});
	}
	if(n==m+1) return cout << (a[1].empty()?"Yes":"No"), 0;
	dsu d;
	d.init(n);
	for(auto i : a[0]) d.unite(i.first, i.second);
	for(auto i : a[1]) if(d.con(i.first, i.second)) quit();
	ll M = n - d.k + d.k*1ll*(d.k-1)/2;
	cout << (m<=M?"Yes":"No");
}