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:
/*
 * Author       : YangDavid
 * Created Time : 2019年09月22日 星期日 20时09分45秒
 */

#include<bits/stdc++.h>
#define rep(i, n) for(int i = 1; i <= n; ++i)
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;

const int maxn = 102000;
int n, q, fa[maxn];
ll m, cc;
vector<pii> op[2];
int find(int v) { return fa[v]==v ? v : fa[v] = find(fa[v]); }
void merge(int x, int y) {
    x = find(x), y = find(y);
    if(x != y) fa[x] = y, cc--;
}
int main() {
    int opt, x, y;
    scanf("%d%lld%d", &n, &m, &q), cc = n;
    rep(i, n) fa[i] = i;
    rep(i, q) {
        scanf("%d%d%d", &x, &y, &opt);
        op[opt].emplace_back(x, y);
    }
    for(auto g : op[0]) merge(g.first, g.second);
    if(m == n - 1) {
        puts(op[1].size() ? "No" : "Yes");
        return 0;
    }

    for(auto g : op[1])
        if(find(g.first) == find(g.second)) { puts("No"); return 0; }
    puts(m <= n-cc + cc*(cc-1)/2 ? "Yes" : "No");
    return 0;
}