Maximum of Maximums of Minimums CodeForces - 872B（思维）

Maximum of Maximums of Minimums

CodeForces - 872B

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.

Examples

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.

题意， 给出一个长度为n的数组， 将其分为连续的k个子集， 为k个子集中最大的最小值是多少？（求出每个子集的最小值， 再求出k个最小值中的最大值）；
当k==1时， 只有原序列， 输出最小值；
当k==2时， 输出a[0], a[n-1]中最大值；

当k>2时， 输出最大值；

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int INF = 0x7fffffff;
const int maxn = 1e5+10;
int main(){
    int n,k;
    cin >> n >> k;
    int a[maxn];
    int maxv = -INF,minv = INF;
    for(int i = 1; i <= n; i++){
        cin >> a[i];
        maxv = max(a[i],maxv);
        minv = min(a[i],minv);
    }
    if(k == 1) printf("%d\n",minv);
    if(k == 2) printf("%d\n",max(a[1],a[n]));
    if(k > 2) printf("%d\n",maxv);
    return 0;
}