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;

#define N 100100

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

int find(int u) {
    return fa[u] == u ? u : fa[u] = find(fa[u]);
}

int main() {
    scanf("%d %lld %d", &n, &m, &q);
    if (m == n - 1) {
        while (q --) {
            int a, b, c;
            scanf("%d %d %d", &a, &b, &c);
            if (c) return puts("No"), 0;
        }
        return puts("Yes"), 0;
    }
    for (int i = 0; i < n; i ++) fa[i] = i;
    for (int i = 0; i < q; i ++) {
        scanf("%d %d %d", &a[i], &b[i], &c[i]);
        if (!c[i]) {
            int u = find(a[i]), v = find(b[i]);
            fa[u] = v;
        }
    }
    for (int i = 0; i < q; i ++) if (c[i] && find(a[i]) == find(b[i])) return puts("No"), 0;
    int k = 0;
    for (int i = 0; i < n; i ++) if (fa[i] == i) k ++;
    if (m <= n + 1ll * k * (k - 3) / 2) puts("Yes");
    else puts("No");

    return 0;
}