CodeForces 574D Bear and Blocks

Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of h_i identical blocks. For clarification see picture for the first sample.

Limak will repeat the following operation till everything is destroyed.

Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.

Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.

Input

The first line contains single integer n (1 ≤ n ≤ 10⁵).

The second line contains n space-separated integers h₁, h₂, ..., h_n (1 ≤ h_i ≤ 10⁹) — sizes of towers.

Output

Print the number of operations needed to destroy all towers.

Examples

Input

6
2 1 4 6 2 2

Output

3

Input

7
3 3 3 1 3 3 3

Output

2

Note

The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color.

After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation.

OJ-ID：
CodeForce 574D

author：
Caution_X

date of submission：
20191019

tags：
dp

description modelling：
给定一个有小正方形组成的不规则图形，现在进行操作:每次都消去暴露在外面的小正方形，问需要几次操作才能消去所有小正方形？

major steps to solve it：
1.dp1[i]:=以第i列为最后一列从前往后可以得到的连续上升子序列
2.dp2[i]:=以第i列为最后一列从后往前可以得到的连续上升子序列
备注：此处连续上升子序列是指可以找到排列成阶梯状的连续上升格子，例如小正方形排列为3 3 3 ，此时仍有连续上升子序列1(3) 2(3) 3
3.对每一列dp取min（dp1,dp2），ans=max(dp[i])

AC code:

#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
int dp1[100005],dp2[100005],a[100005];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++) {
        scanf("%d",&a[i]);
    }
    dp1[n-1]=dp2[n-1]=dp1[0]=dp2[0]=1;
    for(int i=1;i<n;i++) {
        dp1[i]=1;
        if(a[i]>a[i-1]) {
            dp1[i]=dp1[i-1]+1;
        }
        else  {
            if(dp1[i-1]+1<=a[i]) 
                dp1[i]=dp1[i-1]+1;
            else dp1[i]=a[i];
        }
    }
    for(int i=n-2;i>=0;i--) {
        dp2[i]=1;
        if(a[i]>a[i+1]) {
            dp2[i]=dp2[i+1]+1;
        }
        else {
            if(dp2[i+1]+1<=a[i])
                dp2[i]=dp2[i+1]+1;
            else dp2[i]=a[i];
        }
    }
//    for(int i=0;i<n;i++) cout<<dp1[i]<<' ';
//    cout<<endl;
//    for(int i=0;i<n;i++) cout<<dp2[i]<<' ';
//    cout<<endl;
    int ans=0;
    for(int i=0;i<n;i++){
        ans=max(ans,min(dp1[i],dp2[i]));
    }
    printf("%d\n",ans);
    return 0;
}