#include <bits/stdc++.h>

using namespace std;

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

#define rep(pos, len) for(int pos=0;pos<len;pos++)
#define repp(pos, len) for(int pos=1;pos<=len;pos++)

#define INF 987654321
#define IINF 2987654321987654321
#define MOD 1000000007

const int MAXN = 1e5 + 50;
int pa[MAXN];
ll cnt, E;
void init(ll sz) {
	for(int i = 0; i < sz; i++)
		pa[i] = i;
	cnt = sz;
}

int find(int x) {
	return (pa[x] == x) ? x : pa[x] = find(pa[x]);
}

void merge(int x, int y) {
	if(find(x) != find(y)) {
		pa[pa[x]] = pa[y];
		cnt--;
		E++;
	}
}


void yes() {
	printf("Yes\n");
	exit(0);
}

void no() {
	printf("No\n");
	exit(0);
}


bool vis[MAXN];
int main() {
	ll n, m, q;
	scanf("%lld%lld%lld", &n, &m, &q);
	init(n);
	vector<pii> qlist;
	while(q--) {
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c);
		if(c == 0) merge(a, b);
		else qlist.emplace_back(a, b);
	}

	for(auto e : qlist)
		if(find(e.first) == find(e.second))
			no();

	if(E + cnt*(cnt-1)/2 < m) no();
	else if(qlist.empty()) yes();
	else if(cnt <= 2) no();

	for(auto e : qlist)
		vis[find(e.first)] = vis[find(e.second)] = true;

	int edgeNum = 0;
	for(int i = 0; i < n; i++)
		if(vis[i]) edgeNum++;
	edgeNum = max(edgeNum, 3);
	if(m-E >= edgeNum) yes();
	else 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;
typedef pair<ll, ll> pll;

#define rep(pos, len) for(int pos=0;pos<len;pos++)
#define repp(pos, len) for(int pos=1;pos<=len;pos++)

#define INF 987654321
#define IINF 2987654321987654321
#define MOD 1000000007

const int MAXN = 1e5 + 50;
int pa[MAXN];
ll cnt, E;
void init(ll sz) {
	for(int i = 0; i < sz; i++)
		pa[i] = i;
	cnt = sz;
}

int find(int x) {
	return (pa[x] == x) ? x : pa[x] = find(pa[x]);
}

void merge(int x, int y) {
	if(find(x) != find(y)) {
		pa[pa[x]] = pa[y];
		cnt--;
		E++;
	}
}


void yes() {
	printf("Yes\n");
	exit(0);
}

void no() {
	printf("No\n");
	exit(0);
}


bool vis[MAXN];
int main() {
	ll n, m, q;
	scanf("%lld%lld%lld", &n, &m, &q);
	init(n);
	vector<pii> qlist;
	while(q--) {
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c);
		if(c == 0) merge(a, b);
		else qlist.emplace_back(a, b);
	}

	for(auto e : qlist)
		if(find(e.first) == find(e.second))
			no();

	if(E + cnt*(cnt-1)/2 < m) no();
	else if(qlist.empty()) yes();
	else if(cnt <= 2) no();

	for(auto e : qlist)
		vis[find(e.first)] = vis[find(e.second)] = true;

	int edgeNum = 0;
	for(int i = 0; i < n; i++)
		if(vis[i]) edgeNum++;
	edgeNum = max(edgeNum, 3);
	if(m-E >= edgeNum) yes();
	else no();
}
```