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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 100005;

int n, m, q;
int a[N], b[N], c[N];

int p[N];

int fi(int x)
{
    if (x == p[x])
        return x;
    return p[x] = fi(p[x]);
}

void kpc(int x, int y)
{
    x = fi(x);
    y = fi(y);
    p[x] = y;
}

bool solv()
{
    if (m == n - 1)
    {
        for (int i = 0; i < q; ++i)
        {
            if (c[i] == 1)
                return false;
        }
        return true;
    }
    for (int i = 0; i < n; ++i)
        p[i] = i;
    for (int i = 0; i < q; ++i)
    {
        if (c[i] == 0)
            kpc(a[i], b[i]);
    }
    for (int i = 0; i < q; ++i)
    {
        if (c[i] == 1)
        {
            if (fi(a[i]) == fi(b[i]))
                return false;
        }
    }
    set<int> s;
    for (int i = 0; i < n; ++i)
        s.insert(fi(i));
    return (m <= s.size() * 1LL * (s.size() - 1) / 2 + n - s.size());
}

int main()
{
    scanf("%d%d%d", &n, &m, &q);
    for (int i = 0; i < q; ++i)
    {
        scanf("%d%d%d", &a[i], &b[i], &c[i]);
    }
    if (solv())
        printf("Yes\n");
    else
        printf("No\n");
    return 0;
}