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 maxn = 100010;
int n, q, cnt, op[maxn], u[maxn], v[maxn], fa[maxn];
long long m;

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

int main() {
    scanf("%d %lld %d", &n, &m, &q), cnt = n;
    for (int i = 0; i < n; i++) {
        fa[i] = i;
    }
    bool flag = 0;
    for (int i = 1; i <= q; i++) {
        scanf("%d %d %d", &u[i], &v[i], &op[i]);
        if (!op[i]) {
            u[i] = find(u[i]), v[i] = find(v[i]);
            if (u[i] == v[i]) continue;
            fa[u[i]] = v[i], cnt--, m--;
        } else {
            flag = 1;
        }
    }
    if (m == n - 1 && flag) {
        printf("No\n"), exit(0);
    }
    for (int i = 1; i <= q; i++) {
        if (op[i] && find(u[i]) == find(v[i])) {
            printf("No\n"), exit(0);
        }
    }
    if (cnt <= 2 && flag) {
        printf("No\n"), exit(0);
    }
    if (m > 1LL * cnt * (cnt - 1) / 2) {
        printf("No\n"), exit(0);
    }
    printf("Yes\n");
    return 0;
}