Problem:
Score : 700 points

Problem Statement

We have a sequence of N \times K integers: X=(X_0,X_1,\cdots,X_{N \times K-1}).
Its elements are represented by another sequence of N integers: A=(A_0,A_1,\cdots,A_{N-1}). For each pair i, j (0 \leq i \leq K-1,\ 0 \leq j \leq N-1), X_{i \times N + j}=A_j holds.

Snuke has an integer sequence s, which is initially empty.
For each i=0,1,2,\cdots,N \times K-1, in this order, he will perform the following operation:

If s does not contain X_i: add X_i to the end of s.

If s does contain X_i: repeatedly delete the element at the end of s until s no longer contains X_i. Note that, in this case, we do not add X_i to the end of s.

Find the elements of s after Snuke finished the operations.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq K \leq 10^{12}

1 \leq A_i \leq 2 \times 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
A_0 A_1 \cdots A_{N-1}

Output

Print the elements of s after Snuke finished the operations, in order from beginning to end, with spaces in between.

Sample Input 1

3 2
1 2 3

Sample Output 1

2 3

In this case, X=(1,2,3,1,2,3).
We will perform the operations as follows:

i=0: s does not contain 1, so we add 1 to the end of s, resulting in s=(1).

i=1: s does not contain 2, so we add 2 to the end of s, resulting in s=(1,2).

i=2: s does not contain 3, so we add 3 to the end of s, resulting in s=(1,2,3).

i=3: s does contain 1, so we repeatedly delete the element at the end of s as long as s contains 1, which causes the following changes to s: (1,2,3)→(1,2)→(1)→().

i=4: s does not contain 2, so we add 2 to the end of s, resulting in s=(2).

i=5: s does not contain 3, so we add 3 to the end of s, resulting in s=(2,3).

Sample Input 2

5 10
1 2 3 2 3

Sample Output 2

3

Sample Input 3

6 1000000000000
1 1 2 2 3 3

Sample Output 3

s may be empty in the end.

Sample Input 4

11 97
3 1 4 1 5 9 2 6 5 3 5

Sample Output 4

9 2 6

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

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#ifdef tabr
#include "library/debug.cpp"
#else
#define debug(...)
#endif

template <class T>
struct fenwick {
    vector<T> node;
    int n;
    fenwick(int _n)
        : n(_n) {
        node.resize(n);
    }
    void add(int x, T v) {
        while (x < n) {
            node[x] += v;
            x |= (x + 1);
        }
    }
    T get(int x) {
        T v = 0;
        while (x >= 0) {
            v += node[x];
            x = (x & (x + 1)) - 1;
        }
        return v;
    }
    T get(int a, int b) {  // [a, b]
        return get(b) - (a ? get(a - 1) : 0);
    }
    int lower_bound(T v) {
        int x = 0;
        int h = 1;
        while (n >= (h << 1)) {
            h <<= 1;
        }
        for (int k = h; k > 0; k >>= 1) {
            if (x + k <= n && node[x + k - 1] < v) {
                v -= node[x + k - 1];
                x += k;
            }
        }
        return x;
    }
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll n, k;
    cin >> n >> k;
    vector<int> a(n);
    map<int, vector<int>> mp;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        mp[a[i]].emplace_back(i);
    }
    vector<int> order(1);
    vector<int> p(1);
    int cur = 0;
    while (cur != n) {
        int nxt = upper_bound(mp[a[cur]].begin(), mp[a[cur]].end(), cur) - mp[a[cur]].begin();
        p.emplace_back(p.back());
        if (nxt == mp[a[cur]].size()) {
            nxt = 0;
            p.back()++;
        }
        cur = mp[a[cur]][nxt] + 1;
        order.emplace_back(cur);
    }
    p.back()++;
    debug(order);
    debug(p);
    k %= p.back();
    if (k == 0) {
        return 0;
    }
    cur = lower_bound(p.begin(), p.end(), k) - p.begin();
    cur--;
    int pos = order[cur];
    vector<int> ans;
    set<int> st;
    for (int i = pos; i < n; i++) {
        if (st.count(a[i])) {
            while (ans.back() != a[i]) {
                st.erase(ans.back());
                ans.pop_back();
            }
            st.erase(a[i]);
            ans.pop_back();
        } else {
            st.emplace(a[i]);
            ans.emplace_back(a[i]);
        }
    }
    for (int i : ans) {
        cout << i << " ";
    }
    cout << '\n';
    return 0;
}