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 <cstdio>

using namespace std;

int a[100000];
int b[100000];
int c[100000];
int par[100000];

int find(int x) {
    if (par[x] == x) return x;
    return par[x] = find(par[x]);
}

void unite(int x, int y) {
    x = find(x);
    y = find(y);
    
    if (x == y) return;
    
    par[x] = y;
}

int main() {
    int n, q, cnt = 0, i;
    long long m;
    
    scanf("%d %lld %d", &n, &m, &q);
    
    for (i = 0; i < q; i++) scanf("%d %d %d", &a[i], &b[i], &c[i]);
    
    for (i = 0; i < n; i++) par[i] = i;
    
    for (i = 0; i < q; i++) {
        if (c[i] == 1) continue;
        
        unite(a[i], b[i]);
    }
    
    for (i = 0; i < q; i++) {
        if (c[i] == 0) continue;
        
        if (m == n - 1 || find(a[i]) == find(b[i])) {
            puts("No");
            
            return 0;
        }
    }
    
    for (i = 0; i < n; i++) {
        if (par[i] == i) cnt++;
    }
    
    if (m == n - 1 || (cnt > 2 && n + (long long)cnt * (cnt - 3) / 2 >= m)) {
        puts("Yes");
    } else {
        puts("No");
    }
    
    return 0;
}