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;
typedef long long LL;
const int N = 100010;
int fa[N], A[N], B[N], C[N];
int find(int x) {return x == fa[x] ? x : fa[x] = find(fa[x]);}
inline void merge(int x, int y) {fa[find(x)] = find(y);}
int main() {
    int n, q, flg = 0, ct = 0; LL m, L, R = 0; cin >> n >> m >> q;
    for (int i = 1; i <= n; i++) fa[i] = i;
    for (int i = 1; i <= q; i++) {
        cin >> A[i] >> B[i] >> C[i], A[i]++, B[i]++;
        if (!C[i]) merge(A[i], B[i]);
        else flg = 1;
    }
    for (int i = 1; i <= q; i++) if (C[i] && find(A[i]) == find(B[i])) {
        puts("No"); return 0;
    }
    for (int i = 1; i <= n; i++) if (fa[i] == i) ct++;
    L = n - (!flg), R = n - ct + (LL)ct * (ct - 1) / 2;
    if (m < L || m > R) puts("No");
    else puts("Yes");
    return 0;
}