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;
using Int = long long;
struct UnionFind {
    vector<int> A;
    UnionFind(int n) : A(n, -1) {}
    int find(int x) {
        if (A[x] < 0) return x;
        return A[x] = find(A[x]);
    }
    void unite(int x, int y) {
        x = find(x), y = find(y);
        if (x == y) return;
        if (A[x] > A[y]) swap(x, y);
        A[x] += A[y];
        A[y] = x;
    }
};
void okay() { cout << "Yes\n"; exit(0); }
void fail() { cout << "No\n"; exit(0); }
int main() {
    Int N, M, Q; cin >> N >> M >> Q;
    vector<int> A(Q), B(Q), C(Q);
    for (int i = 0; i < Q; i++) cin >> A[i] >> B[i] >> C[i];
    if (M == N - 1) {
        for (int i = 0; i < Q; i++) if (C[i] == 1) fail();
        okay();
    }
    UnionFind uf(N);
    for (int i = 0; i < Q; i++) if (C[i] == 0) {
        uf.unite(A[i], B[i]);
    }
    for (int i = 0; i < Q; i++) if (C[i] == 1) {
        if (uf.find(A[i]) == uf.find(B[i])) {
            fail();
        }
    }
    Int k = 0;
    for (int i = 0; i < N; i++) {
        if (uf.A[i] < 0) k++;
    }
    if (M <= N + k * (k - 3) / 2) {
        okay();
    } else {
        fail();
    }
    return 0;
}