Codeforces Round #210 (Div. 1) B Two points + dp

B. Levko and Array

 

standard output

Levko has an array that consists of integers: a1, a2, ... , an. But he doesn’t like this array at all.

Levko thinks that the beauty of the array a directly depends on value c(a), which can be calculated by the formula:

The less value  c(a) is, the more beautiful the array is.

It’s time to change the world and Levko is going to change his array for the better. To be exact, Levko wants to change the values of at most k array elements (it is allowed to replace the values by any integers). Of course, the changes should make the array as beautiful as possible.

Help Levko and calculate what minimum number c(a) he can reach.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 2000). The second line contains space-separated integers a1, a2, ... , an( - 109 ≤ ai ≤ 109).

Output

A single number — the minimum value of c(a) Levko can get.

Examples

input

Copy

5 2
4 7 4 7 4

output

Copy

0

input

Copy

3 1
-100 0 100

output

Copy

100

input

Copy

6 3
1 2 3 7 8 9

output

Copy

1

Note

In the first sample Levko can change the second and fourth elements and get array: 4, 4, 4, 4, 4.

In the third sample he can get array: 1, 2, 3, 4, 5, 6.

 

Dichotomous + DP problem

ok(x) indicates whether the number of modifications when max(abs(a[i]-a[i+1])) = x is less than or equal to K

ans[i] indicates how much the first i number is modified at least so that the difference is less than x.
When a[i]-a[j] <= x(ij), it means that the i-th and j-th ones can not be modified, and the middle of ji can be modified. The number can make the
difference less than x.

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 const int N = 2020;
 5 ll a[N], ans[N];
 6 ll n, k;
 7 bool ok(ll x) {
 8     ll res = N;
 9     for(ll i = 1; i <= n; i ++) {
10         ans[i] = i-1;
11         for(ll j = 1; j < i; j ++) {
12             if(abs(a[i]-a[j]) <= x*(i-j))
13                 ans[i] = min(ans[i], ans[j]+i-j-1);
14         }
15         res = min(res,ans[i]+n-i);
16     }
17     return res <= k;
18 }
19 int main() {
20     cin >> n >> k;
21     for(int i = 1; i <= n; i ++) cin >> a[i];
22     ll l = 0, r = 2e9;
23     while(l < r) {
24         ll m = (l+r) >> 1;
25         if(ok(m)) r = m;
26         else l = m+1;
27     }
28     printf("%lld\n",r);
29     return 0;
30 }