#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;

#define N 100010
int n, q;
ll m;
vector<pii> f;
int rt[N];

int findrt(int x) {
	if (rt[x] != x) rt[x] = findrt(rt[x]);
	return rt[x];
}

int main() {
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin>>n>>m>>q;
	for (int i = 0; i < n; i ++) rt[i] = i;
	while (q --) {
		int a, b, c;
		cin>>a>>b>>c;
		if (c) {
			f.emplace_back(a, b);
		}
		else {
			int x = findrt(a), y = findrt(b);
			if (x != y) rt[x] = y;
		}
	}
	if (m == n - 1) {
		cout<<(f.empty()?"Yes":"No");
		return 0;
	}
    for (auto &p:f) {
		if (findrt(p.first) == findrt(p.second)) {
			cout<<"No";
			return 0;
		}
    }
    int k = 0;
    for (int i = 0; i < n; i ++) if (findrt(i) == i) k ++;
    cout<<((m <= n + 1ll * k * (k - 3) / 2)?"Yes":"No");
}

### Prompt

Construct a CPP code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;

#define N 100010
int n, q;
ll m;
vector<pii> f;
int rt[N];

int findrt(int x) {
	if (rt[x] != x) rt[x] = findrt(rt[x]);
	return rt[x];
}

int main() {
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin>>n>>m>>q;
	for (int i = 0; i < n; i ++) rt[i] = i;
	while (q --) {
		int a, b, c;
		cin>>a>>b>>c;
		if (c) {
			f.emplace_back(a, b);
		}
		else {
			int x = findrt(a), y = findrt(b);
			if (x != y) rt[x] = y;
		}
	}
	if (m == n - 1) {
		cout<<(f.empty()?"Yes":"No");
		return 0;
	}
    for (auto &p:f) {
		if (findrt(p.first) == findrt(p.second)) {
			cout<<"No";
			return 0;
		}
    }
    int k = 0;
    for (int i = 0; i < n; i ++) if (findrt(i) == i) k ++;
    cout<<((m <= n + 1ll * k * (k - 3) / 2)?"Yes":"No");
}

```