#include <iostream>
#include <map>
#include <set>

const int64_t MAX = 100005;

int64_t N[MAX];
int64_t L[MAX];
struct Query
{
    int64_t a,b,c;
} Q[MAX];
int64_t n,m,q;

int64_t Find(int64_t a)
{
    if (N[a] == a) return a;
    else return N[a] = Find(N[a]);
}

void Union(int64_t l, int64_t r)
{
    if (L[l] < L[r]) Union(r, l);
    else {
        ++L[l];
        N[r] = l;
    }
}

int main()
{
    std::ios_base::sync_with_stdio(0);
    std::cin >> n >> m >> q;
    for (int64_t i=0;i<n;++i) N[i] = i;
    for (int64_t i=0;i<q;++i) std::cin >> Q[i].a >> Q[i].b >> Q[i].c;
    bool result = true;
    bool any = false;
    for (int64_t i=0;i<q;++i)
    {
        if (Q[i].c == 0)
        {
            Union(Find(Q[i].a), Find(Q[i].b));
        }
        else any = true;
    }
    for (int64_t i=0;i<q;++i)
    {   
        if (Q[i].c == 1)
        {
            result &= Find(Q[i].a) != Find(Q[i].b);
        }
    }
    std::map<int64_t, int64_t> C;
    for (int64_t i=0;i<n;++i) C[Find(i)]++;
    int64_t nn = C.size();
    int64_t r = nn*(nn-1)/2;;
    for (auto it : C)
    {
        if (it.second >= 2) r += it.second-1;
    }
    int64_t l = any ? n : n-1;
    result &= l <= m && m <= r;
    std::cout << (result ? "Yes" : "No") << std::endl;

    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 <iostream>
#include <map>
#include <set>

const int64_t MAX = 100005;

int64_t N[MAX];
int64_t L[MAX];
struct Query
{
    int64_t a,b,c;
} Q[MAX];
int64_t n,m,q;

int64_t Find(int64_t a)
{
    if (N[a] == a) return a;
    else return N[a] = Find(N[a]);
}

void Union(int64_t l, int64_t r)
{
    if (L[l] < L[r]) Union(r, l);
    else {
        ++L[l];
        N[r] = l;
    }
}

int main()
{
    std::ios_base::sync_with_stdio(0);
    std::cin >> n >> m >> q;
    for (int64_t i=0;i<n;++i) N[i] = i;
    for (int64_t i=0;i<q;++i) std::cin >> Q[i].a >> Q[i].b >> Q[i].c;
    bool result = true;
    bool any = false;
    for (int64_t i=0;i<q;++i)
    {
        if (Q[i].c == 0)
        {
            Union(Find(Q[i].a), Find(Q[i].b));
        }
        else any = true;
    }
    for (int64_t i=0;i<q;++i)
    {   
        if (Q[i].c == 1)
        {
            result &= Find(Q[i].a) != Find(Q[i].b);
        }
    }
    std::map<int64_t, int64_t> C;
    for (int64_t i=0;i<n;++i) C[Find(i)]++;
    int64_t nn = C.size();
    int64_t r = nn*(nn-1)/2;;
    for (auto it : C)
    {
        if (it.second >= 2) r += it.second-1;
    }
    int64_t l = any ? n : n-1;
    result &= l <= m && m <= r;
    std::cout << (result ? "Yes" : "No") << std::endl;

    return 0;
}
```