#include <bits/stdc++.h>
#define rep(i, x, y) for (int i = x; i <= y; i++)
using namespace std;

const int N = 2e5 + 10;
typedef long long ll;
ll n, m, Q, fa[N], A[N], B[N], tot, num;

int getfa(int x) { return fa[x] == x ? x : fa[x] = getfa(fa[x]); }

int main() {
    cin >> n >> m >> Q;
    rep(i, 0, n - 1) fa[i] = i;
    num = n;  // 连通块个数
    rep(i, 1, Q) {
        int a, b, c; scanf("%d%d%d", &a, &b, &c);
        int x = getfa(a), y = getfa(b);
        if (!c && x != y) fa[y] = x, num--;
        if (c) A[++tot] = a, B[tot] = b;
    }
    rep(i, 1, tot) if (getfa(A[i]) == getfa(B[i])) { puts("No"); return 0; }
    if ((1ll * num * (num - 1) / 2 < m + num - n) || ((num == 2 || m < n) && tot)) puts("No");
    else puts("Yes");
    return 0;
}

### Prompt

Please create a solution in Cpp 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 rep(i, x, y) for (int i = x; i <= y; i++)
using namespace std;

const int N = 2e5 + 10;
typedef long long ll;
ll n, m, Q, fa[N], A[N], B[N], tot, num;

int getfa(int x) { return fa[x] == x ? x : fa[x] = getfa(fa[x]); }

int main() {
    cin >> n >> m >> Q;
    rep(i, 0, n - 1) fa[i] = i;
    num = n;  // 连通块个数
    rep(i, 1, Q) {
        int a, b, c; scanf("%d%d%d", &a, &b, &c);
        int x = getfa(a), y = getfa(b);
        if (!c && x != y) fa[y] = x, num--;
        if (c) A[++tot] = a, B[tot] = b;
    }
    rep(i, 1, tot) if (getfa(A[i]) == getfa(B[i])) { puts("No"); return 0; }
    if ((1ll * num * (num - 1) / 2 < m + num - n) || ((num == 2 || m < n) && tot)) puts("No");
    else puts("Yes");
    return 0;
}

```