CodeForces

You are given a line of nn colored squares in a row, numbered from 11 to nn from left to right. The ii-th square initially has the color cici.

Let's say, that two squares ii and jj belong to the same connected component if ci=cjci=cj, and ci=ckci=ck for all kk satisfying i<k<ji<k<j. In other words, all squares on the segment from ii to jj should have the same color.

For example, the line [3,3,3][3,3,3] has 11 connected component, while the line [5,2,4,4][5,2,4,4] has 33 connected components.

The game "flood fill" is played on the given line as follows:

• At the start of the game you pick any starting square (this is not counted as a turn).
• Then, in each game turn, change the color of the connected component containing the starting square to any other color.

Find the minimum number of turns needed for the entire line to be changed into a single color.

Input

The first line contains a single integer nn (1≤n≤50001≤n≤5000) — the number of squares.

The second line contains integers c1,c2,…,cnc1,c2,…,cn (1≤ci≤50001≤ci≤5000) — the initial colors of the squares.

Output

Print a single integer — the minimum number of the turns needed.

Examples

Input

4
5 2 2 1

Output

2

Input

8
4 5 2 2 1 3 5 5

Output

4

Input

1
4

Output

0
能用区间dp一个很重要的原因是只能通过一个起点更新
代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<cmath>
const int maxn=5e3+5;
typedef long long ll;
using namespace std;
vector<int>vec;
int dp[5005][5005];

int main()
{
    int n;
    cin>>n;
    int x;
    for(int t=1;t<=n;t++)
    {
        scanf("%d",&x);
        vec.push_back(x);
    }
    vec.erase(unique(vec.begin(),vec.end()),vec.end());
    int nn=vec.size();
    memset(dp,0x3f3f3f3f,sizeof(dp));
    for(int t=0;t<nn;t++)
    {
        dp[t][t]=0;
    }
    for(int t=nn-1;t>=0;t--)
    {
        for(int len=1;t+len<nn;len++)
        {
            if(vec[t]==vec[t+len])
            {
                if(len==1)
                {
                    dp[t][t+len]=0;
                }
                else
                {
                    dp[t][t+len]=dp[t+1][t+len-1]+1;
                }
            }
            else
            {
                dp[t][t+len]=min(dp[t][t+len-1],dp[t+1][t+len])+1;
            }
        }
    }
    cout<<dp[0][nn-1]<<endl;
    //system("pause");
    return 0;
}