#include <bits/stdc++.h>
typedef long long i64;
const int N = 1e5+5;
int n, q, flg, cnt, c[N], a[N], b[N], fa[N]; i64 m;
int fd(int x) { return x == fa[x]?x:fa[x] = fd(fa[x]); }
int main()
{
    scanf("%d%lld%d", &n, &m, &q), cnt = n;
    for(int i = 1; i <= n; ++i) fa[i] = i;
    for(int i = 1; i <= q; ++i) scanf("%d%d%d", a+i, b+i, c+i), ++a[i], ++b[i], flg |= c[i];
    for(int i = 1; i <= q; ++i) if(!c[i]&&fd(a[i]) != fd(b[i])) fa[fd(a[i])] = fd(b[i]), --cnt;
    if(m < n-1) return puts("No"), 0;
    if(m == n-1) return puts(!flg?"Yes":"No"), 0;
    for(int i = 1; i <= q; ++i) if(c[i]&&fd(a[i]) == fd(b[i])) return puts("No"), 0;
    puts(m <= n-cnt+(cnt-1ll)*cnt/2?"Yes":"No");
    return 0;
}
### Prompt

Your challenge is to write 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>
typedef long long i64;
const int N = 1e5+5;
int n, q, flg, cnt, c[N], a[N], b[N], fa[N]; i64 m;
int fd(int x) { return x == fa[x]?x:fa[x] = fd(fa[x]); }
int main()
{
    scanf("%d%lld%d", &n, &m, &q), cnt = n;
    for(int i = 1; i <= n; ++i) fa[i] = i;
    for(int i = 1; i <= q; ++i) scanf("%d%d%d", a+i, b+i, c+i), ++a[i], ++b[i], flg |= c[i];
    for(int i = 1; i <= q; ++i) if(!c[i]&&fd(a[i]) != fd(b[i])) fa[fd(a[i])] = fd(b[i]), --cnt;
    if(m < n-1) return puts("No"), 0;
    if(m == n-1) return puts(!flg?"Yes":"No"), 0;
    for(int i = 1; i <= q; ++i) if(c[i]&&fd(a[i]) == fd(b[i])) return puts("No"), 0;
    puts(m <= n-cnt+(cnt-1ll)*cnt/2?"Yes":"No");
    return 0;
}
```