Codeforces Round #440 div2 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, ..., an 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,  ...,  an ( - 109  ≤  ai ≤  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 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 array a is the sequence al,  al + 1,  ...,  ar.

Splitting of array a of n elements into k subsegments [l1, r1], [l2, r2], ..., [lk, rk] (l1 = 1, rk = n, li = ri - 1 + 1 for all i > 1) is ksequences (al1, ..., ar1), ..., (alk, ..., ark).

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 输入 n，k 接下来有一个序列n个数，让你把这个序列分成连续的k段，求出每一段序列最小值中的最大值。

比赛时写这题特判k==2时开了两个数组存前缀后缀最小值，虽然过了但是太麻烦，没想到直接找最前面和最后面这两个数中最大的那个就好了，因为前n-1个数中一定不会有比前1个数更大的满足题意的数了。k==1的时候输出最小值，k==3时输出最大值（把最大那个单独放在一个序列）。

#include <cstdio>#include <cstring>#include <string>#include <iostream>#include <algorithm>#include <set>#include <queue>#include <stack>#include <vector>#include <cmath>using namespace std;#define ll long long#define inf 1000000005#define mod 10000000007#define mem(a, b) memset(a, b, sizeof(a))#define debug() puts("Problem in this!");#define Maxn 100005int main(){        int n, k, a[Maxn], t_max = -inf, t_min = inf;        scanf("%d%d", &n, &k);        for (int i = 0; i < n; i ++) {                scanf("%d", &a[i]);                if (t_max < a[i]) t_max = a[i];                if (t_min > a[i]) t_min = a[i];        }        if (k == 2)                printf("%d\n", max(a[0], a[n - 1]));        else if (k == 1) {                printf("%d\n", t_min);        }        else {                printf("%d\n", t_max);        }        return 0;}