Problem: You are given an array a of n integers. Define the cost of some array t as follows:

$$$cost(t) = ∑_{x ∈ set(t) } last(x) - first(x),$$$ 

where set(t) is the set of all values in t without repetitions, first(x), and last(x) are the indices of the first and last occurrence of x in t, respectively. In other words, we compute the distance between the first and last occurrences for each distinct element and sum them up.

You need to split the array a into k consecutive segments such that each element of a belongs to exactly one segment and the sum of the cost of individual segments is minimum.

Input

The first line contains two integers n, k (1 ≤ n ≤ 35 000, 1 ≤ k ≤ min(n,100)).

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n).

Output

Output the minimum sum of the cost of individual segments.

Examples

Input


7 2
1 6 6 4 6 6 6


Output


3


Input


7 4
5 5 5 5 2 3 3


Output


1

Note

In the first example, we can divide the array into [1,6,6,4] and [6,6,6]. Cost of [1,6,6,4] will be (1-1) + (3 - 2) + (4-4) = 1 and cost of [6,6,6] will be 3-1 = 2. Total cost would be 1 + 2 = 3.

In the second example, divide the array into [5,5],[5],[5,2,3] and [3]. Total Cost would be 1 + 0 + 0 + 0 = 1.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 35001, inf = 2e9 + 5;
int t[4 * N], lz[4 * N], a[N], lst[N], dp[N], ndp[N];
void set_value(int v, int l, int r, int pos, int w) {
    if (l == r) {
        t[v] = w;
        return;
    }
    int mid = (l + r) / 2;
    if (pos <= mid) set_value(2 * v, l, mid, pos, w);
    else set_value(2 * v + 1, mid + 1, r, pos, w);
    t[v] = min(t[2 * v], t[2 * v + 1]);
}
void push(int v, int l, int r) {
    if (lz[v]) {
        lz[2 * v] += lz[v];
        t[2 * v] += lz[v];
        lz[2 * v + 1] += lz[v];
        t[2 * v + 1] += lz[v];
        lz[v] = 0;
    }
}
void update(int v, int l, int r, int a, int b, int w) {
    if (r < a || l > b)
        return;
    if (a <= l && r <= b) {
        t[v] += w;
        lz[v] += w;
        return;
    }
    int mid = (l + r) / 2;
    push(v, l, r);
    update(2 * v, l, mid, a, b, w);
    update(2 * v + 1, mid + 1, r, a, b, w);
    t[v] = min(t[2 * v], t[2 * v + 1]);
}
void check(int v, int l, int r) {
    if (l == r) {
        cout << "t[" << l << "] = ";
        cout << t[v] << " "; return;
    }
    int mid = (l + r) / 2;
    check(2 * v, l, mid);
    check(2 * v + 1, mid + 1, r);
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int n, k; cin >> n >> k;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    // dp_{i,j} = \min\limits_{k < i}(dp_{k,j - 1} + cost(k + 1, i))
    for (int i = 1; i <= n; i++) {
        dp[i] = inf;
    }
    dp[0] = 0;
    while (k--) {
        for (int i = 0; i < 4 * N; i++) {
            t[i] = inf;
            lz[i] = 0;
        }
        memset(lst, 0, (n + 1) * sizeof *lst);
        for (int i = 1; i <= n; i++) {
            set_value(1, 0, n - 1, i - 1, dp[i - 1]);
            if (lst[a[i]]) {
                update(1, 0, n - 1, 0, lst[a[i]] - 1, i - lst[a[i]]);
            }
            lst[a[i]] = i;
            ndp[i] = t[1];
        }
        for (int i = 0; i <= n; i++)
            dp[i] = ndp[i];
    }
    cout << dp[n] << '\n';
    return 0;
}