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=4e5,M=110,inf=0x3f3f3f3f;
int a[N],dp[N][M],lst[N],pre[N],nxt[N],i,j,n,m,L,R,sum;
int cal(int l,int r){
	while(L<l){if(nxt[L]<=R)sum-=nxt[L]-L;L++;}
	while(L>l){--L;if(nxt[L]<=R)sum+=nxt[L]-L;}
	while(R<r){++R;if(pre[R]>=L)sum+=R-pre[R];}
	while(R>r){if(pre[R]>=L)sum-=R-pre[R];R--;}
	return sum;
} 
void solve(int l,int r,int L,int R,int now){
	if(l>r||L>R)return;int mid=l+r>>1,val=inf,pos;
	for(int i=L;i<mid&&i<=R;++i){
		int tmp=dp[i][now-1]+cal(i+1,mid);
		if(tmp<val)pos=i,val=tmp;
	}dp[mid][now]=val;solve(l,mid-1,L,pos,now);solve(mid+1,r,pos,R,now);
}
int main(){
	for(scanf("%d%d",&n,&m),i=1;i<=n;++i)scanf("%d",a+i);
	for(i=1;i<=n;++i)dp[i][0]=inf,pre[i]=lst[a[i]],lst[a[i]]=i;
	for(i=1;i<=n;++i)lst[a[i]]=n+1;
	for(i=n;i;--i)nxt[i]=lst[a[i]],lst[a[i]]=i;
	for(i=1;i<=m;++i)solve(1,n,0,n,i);printf("%d\n",dp[n][m]);
}