Codeforces Round #440 (Div. 2）B. Maximum of Maximums of Minimums

B. Maximum of Maximums of Minimums

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 thek 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 andk (1 ≤ k ≤ n ≤  10⁵) — the size of the arraya 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 intok non-empty subsegments and take maximum of minimums on the subsegments.

Examples

Input

5 21 2 3 4 5

Output

5

Input

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

Output

-5

Note

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

题目意思：

给你两个数字n和k   把这n个数字分成k段 让这k段中的k个最小值中的最大值最大

思路：

当k==1的时候我们直接输出最小值就好了

当k>2的时候我们可以单独把最大值隔开 然后能找到的最大值就一定是原序列的最大值本身

当k==2的时候  我用了一个前缀最小值数组和一个后缀最小值数组来维护前n项中的最小值和后n项中的最小值

本来这个操作需要一个for来遍历的 但是后来发现这个array[0]  and array[n - 1] 一定是前缀最小值数组和后缀最小值数组中最大的值

所以只需要比较这两个数字的大小输出大的就好~~

比赛的时候我这道题莫名TL4次 就一个for循环 我至今不知道为什么~~~

#include <cstdio>#include <cstring>#include <algorithm>const int inf = 0x7fffffff;const int N = 100005;using namespace std;int array[N];int main(){int n , k;scanf("%d%d",&n,&k);int maxx = -inf;int minn = inf;int max_id = 0;for (int i = 0 ; i < n ; i++) {scanf("%d",&array[i]);if(array[i] > maxx) {maxx = array[i];max_id = i;}minn = min(minn , array[i]);}if(k == 1) {printf("%d\n",minn);} else if (k > 2) {printf("%d\n",maxx);}else {printf("%d\n",max(array[n-1],array[0]));}}