分治

Bizon the Champion isn't just attentive, he also is very hardworking.

Bizon the Champion decided to paint his old fence his favorite color, orange. The fence is represented as n vertical planks, put in a row. Adjacent planks have no gap between them. The planks are numbered from the left to the right starting from one, the i-th plank has the width of 1 meter and the height of a_i meters.

Bizon the Champion bought a brush in the shop, the brush's width is 1 meter. He can make vertical and horizontal strokes with the brush. During a stroke the brush's full surface must touch the fence at all the time (see the samples for the better understanding). What minimum number of strokes should Bizon the Champion do to fully paint the fence? Note that you are allowed to paint the same area of the fence multiple times.

Input

The first line contains integer n (1 ≤ n ≤ 5000) — the number of fence planks. The second line contains n space-separated integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹).

Output

Print a single integer — the minimum number of strokes needed to paint the whole fence.

Example

Input

5
2 2 1 2 1

Output

3

Input

2
2 2

Output

2

Input

1
5

Output

1

Note

In the first sample you need to paint the fence in three strokes with the the first stroke goes on height 1 horizontally along all the planks. The second stroke goes on height 2 horizontally and paints the first and second planks and the third stroke (it can be horizontal and vertical) finishes painting the fourth plank.

In the second sample you can paint the fence with two strokes, either two horizontal or two vertical strokes.

In the third sample there is only one plank that can be painted using a single vertical stroke.

题目分析 ： 给出以堆木板的高度，木板可以横着刷，也可以竖着刷，问最终最小的操作次数是多少？

思路分析 ： 首先对于这些木板最坏的刷法，就是全部都是竖着刷，那么答案就是 n ，另一种就是横着去刷，横着去把木板长度最短的木板去刷了，然后这样呢我们就会得到一组新的木板长度，它们之间可能是被很多长度为 0  的木板分割开来的，那么被分割出来的这些木板我们是不就可以看成最初的初始的状态，（什么样的情况才会分出这些木板呢，是不就是木板最短的板它是一定会被刷掉的）

代码示例 ：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <time.h>
using namespace std;
#define ll long long
const int maxn = 1e6+5;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;

int a[5005];

int dfs(int l, int r){
    int minn = inf;
    int cnt = 0;
    for(int i = l; i <= r; i++){
        if (a[i] < minn) minn = a[i];
    }
    for(int i = l; i <= r; i++){
        a[i] -= minn;
    }
    int p = l;
    int sum = minn;
    for(int i = l; i <= r; i++){
        if (a[i] != 0) {p = i; break;}
    }
    for(int i = p; i <= r; i++){
        if (a[i] == 0) {sum += dfs(p, i-1); p = i+1;}
    }
    if (p <= r) sum += dfs(p, r);
    
    //printf("%d  %d  %d  %d
", l, r, sum, r-l+1);
    return min(sum, r-l+1);
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int n;
    
    cin >> n;
    for(int i = 1; i <= n; i++) cin >> a[i];
    
    printf("%d
", dfs(1, n));
    return 0;
}