#include<bits/stdc++.h>
#define ll long long int
#define f first
#define s second
#define pb push_back
#define PII pair<ll,ll>
ll const MOD = 1000000007;
ll const N = 1000005;
ll const LN = 20;
ll const inf = 8e18;
using namespace std;

ll n, m, a[N];

ll id[N];
ll sz[N];
void ini()
{
	for (ll i = 0; i < N; i++) { id[i] = i, sz[i] = 1; }
}
ll root (ll i)
{
	while (i != id[i]) { id[i] = id[id[i]], i = id[i]; }
	return i;
}
void uni (ll a, ll b)
{
	ll ra = root (a);
	ll rb = root (b);
	if (ra == rb) { return; }
	if (sz[ra] > sz[rb]) { id[rb] = ra, sz[ra] += sz[rb]; }
	else { id[ra] = rb, sz[rb] += sz[ra]; }
}
vector<PII>v[2];
int main()
{
	ios::sync_with_stdio (false);
	cin.tie (NULL);
	ini();
	ll n, m, q;
	cin >> n >> m >> q;
	for (ll i = 1; i <= q; i++)
	{
		ll a, b, c;
		cin >> a >> b >> c;
		a++;
		b++;
		v[c].pb ({a, b});
		if (!c)
		{
			uni (a, b);
		}
	}
	for (auto w : v[1])
	{
		if (root (w.f) == root (w.s))
		{
			cout << "No";
			return 0;
		}
	}
	ll cmp = 0;
	for (ll i = 1; i <= n; i++)
	{
		if (root (i) == i) { cmp++; }
	}
	ll l = n;
	ll r = cmp * (cmp - 1) / 2 + (n - cmp);
	if (v[1].size() == 0)
	{
		l--;
	}
	if (l <= m && m <= r)
	{
		cout << "Yes";
	}
	else { cout << "No"; }
	return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 ll long long int
#define f first
#define s second
#define pb push_back
#define PII pair<ll,ll>
ll const MOD = 1000000007;
ll const N = 1000005;
ll const LN = 20;
ll const inf = 8e18;
using namespace std;

ll n, m, a[N];

ll id[N];
ll sz[N];
void ini()
{
	for (ll i = 0; i < N; i++) { id[i] = i, sz[i] = 1; }
}
ll root (ll i)
{
	while (i != id[i]) { id[i] = id[id[i]], i = id[i]; }
	return i;
}
void uni (ll a, ll b)
{
	ll ra = root (a);
	ll rb = root (b);
	if (ra == rb) { return; }
	if (sz[ra] > sz[rb]) { id[rb] = ra, sz[ra] += sz[rb]; }
	else { id[ra] = rb, sz[rb] += sz[ra]; }
}
vector<PII>v[2];
int main()
{
	ios::sync_with_stdio (false);
	cin.tie (NULL);
	ini();
	ll n, m, q;
	cin >> n >> m >> q;
	for (ll i = 1; i <= q; i++)
	{
		ll a, b, c;
		cin >> a >> b >> c;
		a++;
		b++;
		v[c].pb ({a, b});
		if (!c)
		{
			uni (a, b);
		}
	}
	for (auto w : v[1])
	{
		if (root (w.f) == root (w.s))
		{
			cout << "No";
			return 0;
		}
	}
	ll cmp = 0;
	for (ll i = 1; i <= n; i++)
	{
		if (root (i) == i) { cmp++; }
	}
	ll l = n;
	ll r = cmp * (cmp - 1) / 2 + (n - cmp);
	if (v[1].size() == 0)
	{
		l--;
	}
	if (l <= m && m <= r)
	{
		cout << "Yes";
	}
	else { cout << "No"; }
	return 0;
}

```