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<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=35003;
int a[N],first[N],last[N],pre[N],dp[N];
ll b[N],dat[N << 2], cov[N << 2];
void inline pushup(int p)
{
    dat[p] = min(dat[p << 1], dat[(p << 1) | 1]);
}
void build(int p, int l, int r)
{
    dat[p] = cov[p] = 0;
    if (l == r)
    {
        dat[p] = b[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(p << 1, l, mid);
    build((p << 1) | 1, mid + 1, r);
    pushup(p);
}
void pushdown(int p, int l, int r)
{
    if (cov[p])
    {
        int mid = (l + r) >> 1;
        dat[p << 1] += cov[p];
        cov[p << 1] += cov[p];
        dat[(p << 1) | 1] += cov[p];
        cov[(p << 1) | 1] += cov[p];
        cov[p] = 0;
    }
}
void change(int p, int l, int r, int ql, int qr, int x)
{
    if (ql <= l && r <= qr)
    {
        dat[p] += x;
        cov[p] += x;
        return;
    }
    pushdown(p, l, r);
    int mid = (l + r) >> 1;
    if (ql <= mid)
        change(p << 1, l, mid, ql, qr, x);
    if (qr > mid)
        change((p << 1) | 1, mid + 1, r, ql, qr, x);
    pushup(p);
}
ll ask(int p, int l, int r, int ql, int qr)
{
    if (ql <= l && r <= qr)
        return dat[p];
    pushdown(p, l, r);
    int mid = (l + r) >> 1;
    ll ret = 1e18;
    if (ql <= mid)
        ret = min(ret,ask(p << 1, l, mid, ql, qr));
    if (qr > mid)
        ret = min(ret,ask((p << 1) | 1, mid + 1, r, ql, qr));
    return ret;
}
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for (int i = 1; i <= n; i++)
    {
        scanf("%d",a + i);
        b[i] = b[i-1];
        if (last[a[i]])
            pre[i]=last[a[i]],b[i]+=i-pre[i];
        else
            pre[i]=i,first[a[i]]=i;
        last[a[i]] = i;
    }
    for (int t = 1; t < m; t++)
    {
        build(1,t,n);
        for (int i = t + 1; i <= n; i++)
        {
            if (i!=pre[i]&&pre[i]>t)
                change(1,t,n,t,max(t,pre[i]-1),i-pre[i]);
            b[i]=ask(1,t,n,t,i-1);
            // printf("%lld ",b[i]);
        }
        // printf("\n");
    }
    printf("%lld\n",b[n]);
    return 0;
}