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;

#define Debug(x) x
typedef int regular_int;
#define int ll
typedef long long ll;

vector<int > G[200000];
int cnt[200000];

void magic(int x, int y){
    cnt[x] = y;
    for (int z: G[x]){
        if (!cnt[z]){
            magic(z, y);

        }
    }
}


regular_int main() {
	ios_base::sync_with_stdio(false);
	cout << setprecision(12) << fixed;

	int N, M, Q;
	cin >> N >> M >> Q;
	vector<pair<int, int > > dist;
	for (int i = 0; i < Q; ++i) {
        int A, B, C;
        cin >> A >> B >> C;

        if (C){
            dist.push_back(make_pair(A, B));

        } else {
            G[A].push_back(B);
            G[B].push_back(A);


        }
	}

	int sum = 0;
	for (int i = 0; i < N; ++i) {
        if (!cnt[i]){
            magic(i, ++sum);

        }
	}

	for (auto x: dist){
        if (cnt[x.first] == cnt[x.second]){
            cout << "No" << endl;
            return 0;
        }
	}

	if (!dist.empty() && sum < 3){
        cout << "No" << endl;
        return 0;
	}

	if (M > (ll )sum * (sum - 1) / 2 + N - sum){
        cout << "No" << endl;
        return 0;
	}

	if (!dist.empty() && M < N){
        cout << "No" << endl;
        return 0;
	}

	cout << "Yes" << endl;

	return 0;
}