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;

int main() {
    int n, q; long long m;
    scanf("%d%lld%d", &n, &m, &q);
    vector<int> uf(n);
    for (int i = 0; i < n; ++i) uf[i] = i;

    function<int(int)> Find = [&](int x) {
        if (x == uf[x]) return x;
        return uf[x] = Find(uf[x]);
    };

    vector<pair<int, int>> e;
    for (int i = 0; i < q; ++i) {
        int u, v, c; scanf("%d%d%d", &u, &v, &c);
        if (c == 0) uf[Find(u)] = Find(v);
        else e.emplace_back(u, v);
    }
    for (int i = 0; i < e.size(); ++i) {
        if (Find(e[i].first) == Find(e[i].second)) {
            puts("No");
            return 0;
        }
    }
    int k = 0;
    for (int i = 0; i < n; ++i) if (Find(i) == i) ++k;
    long long km = 1LL * k * (k - 1) / 2 + n - k;
    if (m == n - 1) {
        if (e.empty()) puts("Yes");
        else puts("No");
    } else {
        if (m <= km) puts("Yes");
        else puts("No");
    }
    return 0;
}