#include <bits/stdc++.h>
#define mod 1000000007
#define mod998 998244353
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define mkp make_pair
typedef long long ll;
using namespace std;

int N, Q, A[100000], B[100000], C[100000], t[100000];
bool f;
ll M;

int par(int x) {
	if (t[x] < 0)return x;
	return t[x] = par(t[x]);
}

bool u(int x, int y) {
	x = par(x);
	y = par(y);
	if (x == y)return false;
	if (t[x] < t[y]) {
		t[x] += t[y];
		t[y] = x;
	}
	else {
		t[y] += t[x];
		t[x] = y;
	}
	return true;
}


int main() {
	cin >> N >> M >> Q;
	for (int i = 0; i < Q; ++i) {
		cin >> A[i] >> B[i] >> C[i];
	}
	for (int i = 0; i < N; ++i) {
		t[i] = -1;
	}
	for (int i = 0; i < Q; ++i) {
		if (!C[i]) {
			M -= u(A[i], B[i]);
		}
		else f = true;
	}
	ll c = 0;
	for (int i = 0; i < N; ++i) {
		if (t[i] < 0) {
			++c;
		}
	}
	for (int i = 0; i < Q; ++i) {
		if (C[i] && par(A[i]) == par(B[i])) {
			cout << "No" << endl;
			return 0;
		}
	}

	if (c*(c - 1) / 2 < M) {
		cout << "No" << endl;
		return 0;
	}
	else{
		if (!f)--c;
		if (M < c) {
			cout << "No" << endl;
			return 0;
		}
	}
	cout << "Yes" << endl;
}
### Prompt

Please formulate a cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
#define mod 1000000007
#define mod998 998244353
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define mkp make_pair
typedef long long ll;
using namespace std;

int N, Q, A[100000], B[100000], C[100000], t[100000];
bool f;
ll M;

int par(int x) {
	if (t[x] < 0)return x;
	return t[x] = par(t[x]);
}

bool u(int x, int y) {
	x = par(x);
	y = par(y);
	if (x == y)return false;
	if (t[x] < t[y]) {
		t[x] += t[y];
		t[y] = x;
	}
	else {
		t[y] += t[x];
		t[x] = y;
	}
	return true;
}


int main() {
	cin >> N >> M >> Q;
	for (int i = 0; i < Q; ++i) {
		cin >> A[i] >> B[i] >> C[i];
	}
	for (int i = 0; i < N; ++i) {
		t[i] = -1;
	}
	for (int i = 0; i < Q; ++i) {
		if (!C[i]) {
			M -= u(A[i], B[i]);
		}
		else f = true;
	}
	ll c = 0;
	for (int i = 0; i < N; ++i) {
		if (t[i] < 0) {
			++c;
		}
	}
	for (int i = 0; i < Q; ++i) {
		if (C[i] && par(A[i]) == par(B[i])) {
			cout << "No" << endl;
			return 0;
		}
	}

	if (c*(c - 1) / 2 < M) {
		cout << "No" << endl;
		return 0;
	}
	else{
		if (!f)--c;
		if (M < c) {
			cout << "No" << endl;
			return 0;
		}
	}
	cout << "Yes" << endl;
}
```