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 ll = long long;
const int N = 100010;

int a[N], b[N], c[N], fa[N];

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

void unite(int u, int v){
    u = find(u), v = find(v);
    if (u == v){
        return;
    }
    fa[u] = v;
}

int main(){
    int n, q;
    ll m;
    scanf("%d%lld%d", &n, &m, &q);
    for (int i = 0; i < n; ++ i){
        fa[i] = i;
    }
    bool flag = false;
    for (int i = 0; i < q; ++ i){
        scanf("%d%d%d", &a[i], &b[i], &c[i]);
        if (c[i] == 0){
            unite(a[i], b[i]);
        }
        else{
            flag = true;
        }
    }
    for (int i = 0; i < q; ++ i){
        if (c[i] == 1){
            if (find(a[i]) == find(b[i])){
                puts("No");
                return 0;
            }
        }
    }
    int cnt = 0;
    for (int i = 0; i < n; ++ i){
        if (find(i) == i){
            ++ cnt;
        }
    }
    ll left = n - 1 + int(flag);
    ll right = n - 1 + 1ll * cnt * (cnt - 1) / 2 - (cnt - 1);
    puts(left <= m && m <= right ? "Yes" : "No");
    return 0;
}