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>
#define ll long long
using namespace std;
const int MAXN = 1e5 + 7;
int x[MAXN], y[MAXN], c[MAXN];
int f[MAXN], cnt0, cnt1, used;
int find(int x) {return x == f[x] ? f[x] : f[x] = find(f[x]);}
void unite(int x, int y)
{
    x = find(x), y = find(y);
    if (x != y) f[x] = y, cnt0--, used++;
}
int main()
{
    int n, q; ll m; scanf("%d%lld%d", &n, &m, &q);
    cnt0 = n; used = 0; cnt1 = 0;
    for (int i = 1; i <= n; i++) f[i] = i;
    for (int i = 1; i <= q; i++)
    {
        scanf("%d%d%d", &x[i], &y[i], &c[i]);
        x[i]++; y[i]++;
        if (!c[i]) unite(x[i], y[i]);
        else 
        {
            cnt1++;
            if (m == n - 1) {puts("No"); return 0;}
        }
    }
    m -= used;
    if (!cnt1)
    {
        (m >= cnt0 - 1 && m <= (ll)cnt0 * (cnt0 - 1) / 2) ? puts("Yes") : puts("No");
        return 0;
    }
    else 
    {
        if (cnt0 == 2) {puts("No"); return 0;}
        for (int i = 1; i <= q; i++) if (c[i])
            if (find(x[i]) == find(y[i])) {puts("No"); return 0;}
        (m >= cnt0 && m <= (ll)cnt0 * (cnt0 - 1) / 2) ? puts("Yes") : puts("No");
    }
    return 0;
}