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 Maxn = 100005;

int n;
ll m;
int q;
int par[Maxn], siz[Maxn];
int a[Maxn], b[Maxn], c[Maxn];

int getPar(int x) { return par[x] == x? x: par[x] = getPar(par[x]); }

bool unionSet(int a, int b)
{
    a = getPar(a), b = getPar(b);
    if (a == b) return false;
    if (siz[a] < siz[b]) swap(a, b);
    siz[a] += siz[b]; par[b] = a;
    return true;
}

int main()
{
    scanf("%d %lld %d", &n, &m, &q);
    for (int i = 0; i < n; i++)
        par[i] = i, siz[i] = 1;
    bool was1 = false;
    for (int i = 0; i < q; i++) {
        scanf("%d %d %d", &a[i], &b[i], &c[i]);
        if (c[i] == 0) {
            if (unionSet(a[i], b[i])) m--;
        } else was1 = true;
    }
    for (int i = 0; i < q; i++) if (c[i] == 1)
        if (getPar(a[i]) == getPar(b[i])) { printf("No\n"); return 0; }
    int cnt = 0;
    for (int i = 0; i < n; i++)
        if (getPar(i) == i) cnt++;
    ll lef = ll(was1) + ll(cnt - 1);
    ll rig = ll(cnt) * (cnt - 1) / 2;
    if (lef <= m && m <= rig)
        printf("Yes\n");
    else printf("No\n");
    return 0;
}