C - Maximum of Maximums of Minimums（数学）

C - Maximum of Maximums of Minimums

You are given an array a₁, a₂, ..., a_n consisting of n integers, and an integer k. You have to split the array into exactly k non-empty subsegments. You'll then compute the minimum integer on each subsegment, and take the maximum integer over the k obtained minimums. What is the maximum possible integer you can get?

Definitions of subsegment and array splitting are given in notes.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤  10⁵) — the size of the array a and the number of subsegments you have to split the array to.

The second line contains n integers a₁,  a₂,  ...,  a_n ( - 10⁹  ≤  a_i ≤  10⁹).

Output

Print single integer — the maximum possible integer you can get if you split the array into k non-empty subsegments and take maximum of minimums on the subsegments.

Example

Input

5 2
1 2 3 4 5

Output

5

Input

5 1
-4 -5 -3 -2 -1

Output

-5

Note

A subsegment [l,  r] (l ≤ r) of array a is the sequence a_l,  a_l + 1,  ...,  a_r.

Splitting of array a of n elements into k subsegments [l₁, r₁], [l₂, r₂], ..., [l_k, r_k] (l₁ = 1, r_k = n, l_i = r_i - 1 + 1 for all i > 1) is k sequences (a_l1, ..., a_r1), ..., (a_lk, ..., a_rk).

In the first example you should split the array into subsegments [1, 4] and [5, 5] that results in sequences (1, 2, 3, 4) and (5). The minimums are min(1, 2, 3, 4) = 1 and min(5) = 5. The resulting maximum is max(1, 5) = 5. It is obvious that you can't reach greater result.

In the second example the only option you have is to split the array into one subsegment [1, 5], that results in one sequence ( - 4,  - 5,  - 3,  - 2,  - 1). The only minimum is min( - 4,  - 5,  - 3,  - 2,  - 1) =  - 5. The resulting maximum is  - 5

水题，关键是搞清楚题意，还有当k==2时，为什么是max（a[0], a[n-1]）,为什么a[0],和a[n-1]一定是序列里的最小值？？？？

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>

using namespace std;

int main()
{
    int n,k;
    int i,j;
    int ans;
    int mmax,mmin;
    int a[];
    mmax = -1e9-;
    mmin = 1e9+;
    scanf("%d %d",&n, &k);
    for(i = ; i < n; i++)
    {
        scanf("%d",&a[i]);
    }
    if(k == )
    {
        for(i = ; i < n; i++)
        {
            mmin = min(a[i], mmin);
        }
        printf("%d\n",mmin);
    }
    else if(k == )
    {
        ans = max(a[], a[n-]);
        printf("%d\n",ans);
    }
    else if(k >= )
    {
        for(i = ; i < n; i++)
        {
            mmax = max(a[i], mmax);
        }
        printf("%d\n",mmax);
    }
    return ;
}