Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

5 5 3
0 1 0
1 2 1
2 3 0

Sample Output 1

Yes

For example, consider a graph with edges (0,1),(1,2),(1,4),(2,3),(2,4). This graph matches the clues.

Sample Input 2

4 4 3
0 1 0
1 2 1
2 3 0

Sample Output 2

No

Sample Input 3

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

Sample Output 3

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#include <random>

using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
#define double ld
//#define int ll
//#define int __int128
//#define ll __int128
typedef pair<int, int   > pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector< vi > vvi;
typedef vector< vvi > vvvi;
typedef vector<short> vs;
typedef vector<vs> vvs;
typedef vector<vvs> vvvs;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<vvl> vvvl;
typedef vector<ld> vld;
typedef vector<vld> vvld;
typedef vector<vvld> vvvld;
typedef vector<string> vst;
typedef vector<vst> vvst;
typedef pair<ld, ld> pld;
//typedef complex<double> base;

#define inmin(a, b) a = min(a, (b))
#define inmax(a, b) a = max(a, (b))
#define ALL(a) a.begin(),a.end()
#define RALL(a) a.rbegin(),a.rend()
#define sqr(x) ((x) * (x))
#define fori(i, n) for(int i = 0; i < (int)(n); ++i)
#define SZ(a) ((int)((a).size()))
#define triple(T) tuple<T, T, T>
#define quad(T) tuple<T, T, T, T>
#define watch(x) cerr << (#x) << " = " << (x) << endl;

#ifdef RUS_HOME
#define cerr cout
#else
#define cerr if (false) cerr
#endif

const double PI = 2 * acos(0.0);
#define rand shittttty_shit
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());

const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";

template <class T0, class T1>
inline ostream & operator << (ostream &out, pair<T0, T1> &a) {
    return out << "{" << a.first << ", " << a.second << "}";
}

template <class T0, class T1, class T2>
inline ostream & operator << (ostream &out, tuple<T0, T1, T2> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a) << "}";
}

template <class T0, class T1, class T2, class T3>
inline ostream & operator << (ostream &out, tuple<T0, T1, T2, T3> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a) << ", " <<  get<3>(a) << "}";
}

template<class T>
inline ostream & operator << (ostream &out, vector<T> &a) {
    out << "[";
    fori (i, a.size())
        out << a[i] << vector<string>{", ", "]  "}[i + 1 == a.size()];
    return out;
}



void smain();

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

#ifdef RUS_HOME
    freopen("input.txt", "r", stdin);
    clock_t start = clock();
#endif
    cout << setprecision(12) << fixed;
    smain();
#ifdef RUS_HOME
    cout << "\n\n\n\nTOTAL EXECUTION TIME: " << float( clock () - start ) /  CLOCKS_PER_SEC << endl;
#endif
    return 0;
}

const int N = 1e6 + 1, oo = 1e9, MOD = 998244353;

int pr[N], pr2[N];
vi g[N], g2[N];

int kek, lol;
void merge(int a, int b) {
    a = pr[a]; b = pr[b];
    if (a == b)
        return;
    kek--;
    if (g[a].size() < g[b].size())
        swap(a, b);
    for (auto p : g[b]) {
        pr[p] = a;
        g[a].push_back(p);
    }
    g[b].clear();
}

void merge2(int a, int b) {
    a = pr2[a]; b = pr2[b];
    if (a == b)
        return;
    lol--;
    if (g2[a].size() < g2[b].size())
        swap(a, b);
    for (auto p : g2[b]) {
        pr2[p] = a;
        g2[a].push_back(p);
    }
    g2[b].clear();
}

void smain() {

    int n, m, q;
    cin >> n >> m >> q;
    kek = n;
    for (int i = 0; i < n; i++) {
        pr[i] = i;
        g[i].push_back(i);
    }
    vi a(q), b(q);
    vpi d(q);
    for (int i = 0; i < q; i++) {
        cin >> a[i] >> b[i] >> d[i].first;
        d[i].second = i;
    }
    sort(ALL(d));

    for (int i = 0; i < q; i++) {
        if (d[i].first == 0) {
            merge(a[d[i].second], b[d[i].second]);
        } else
            break;
    }
    vector <bool> used(n);
    for (int i = 0; i < n; i++) {
        int u = pr[i];
        if (!used[u]) {
            pr2[u] = u;
            g2[u].push_back(u);
            used[u] = true;
        }
        if (i != u) {
            pr2[i] = oo;
        }
    }


    lol = kek;
    for (int i = q - 1; i >= 0; i--) {
        if (d[i].first == 1) {
            if (pr[a[d[i].second]] == pr[b[d[i].second]]) {
                cout << "No";
                return;
            } else {
                merge2(pr[a[d[i].second]], pr[b[d[i].second]]);
            }
        } else
            break;
    }

    int ans = n - kek;
    vector <bool> used2(n);
    for (int i = 0; i < n; i++) {
        int y = pr2[i];
        if (y != oo && !used2[y]) {
            used2[y] = true;
            if (g2[y].size() > 2) {
                ans += g2[y].size() - 1;
            } else if (g2[y].size() == 2) {
                if (lol == 1) {
                    cout << "No";
                    return;
                } else {
                    ans += 1;
                }
            }
        }
    }
    if (lol != kek) {
        ans = n;
    } else {
        ans = n - 1;
    }
    if (ans <= m) {
        cout << "Yes";
    } else {
        cout << "No";
    }





}