Problem:
Score : 400 points

Problem Statement

Your friend gave you a dequeue D as a birthday present.

D is a horizontal cylinder that contains a row of N jewels.

The values of the jewels are V_1, V_2, ..., V_N from left to right. There may be jewels with negative values.

In the beginning, you have no jewel in your hands.

You can perform at most K operations on D, chosen from the following, at most K times (possibly zero):

Operation A: Take out the leftmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.

Operation B: Take out the rightmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.

Operation C: Choose a jewel in your hands and insert it to the left end of D. You cannot do this operation when you have no jewel in your hand.

Operation D: Choose a jewel in your hands and insert it to the right end of D. You cannot do this operation when you have no jewel in your hand.

Find the maximum possible sum of the values of jewels in your hands after the operations.

Constraints

All values in input are integers.

1 \leq N \leq 50

1 \leq K \leq 100

-10^7 \leq V_i \leq 10^7

Input

Input is given from Standard Input in the following format:

N K
V_1 V_2 ... V_N

Output

Print the maximum possible sum of the values of jewels in your hands after the operations.

Sample Input 1

6 4
-10 8 2 1 2 6

Sample Output 1

14

After the following sequence of operations, you have two jewels of values 8 and 6 in your hands for a total of 14, which is the maximum result.

Do operation A. You take out the jewel of value -10 from the left end of D.

Do operation B. You take out the jewel of value 6 from the right end of D.

Do operation A. You take out the jewel of value 8 from the left end of D.

Do operation D. You insert the jewel of value -10 to the right end of D.

Sample Input 2

6 4
-6 -100 50 -2 -5 -3

Sample Output 2

44

Sample Input 3

6 3
-6 -100 50 -2 -5 -3

Sample Output 3

0

It is optimal to do no operation.

Solution in JavaScript (Node.js 12.16.1) (judge verdict: Wrong Answer)

//↓見なくていいよ！ここから--------------------------------------------------
var read = require('readline').createInterface({
  input: process.stdin,
  output: process.stdout
});
var obj;
var inLine = [];
read.on('line', function(input){inLine.push(input);});
read.on('close', function(){
  obj = init(inLine);
  console.error("\n");
  var start = Date.now();
  Main();
  var end = Date.now() - start;
  myerr("time : " + (end) + "ms");
  myerr("memory : " + Math.round(process.memoryUsage().heapTotal / 1024) + "KB");
});
function makeClone(obj){return JSON.parse(JSON.stringify(obj));}
function nextInt(){return myconv(next(),1);} function nextStrArray(){return myconv(next(),2);}
function nextIntArray(){return myconv(next(),4);} function nextCharArray(){return myconv(next(),6);}
function next(){return obj.next();} function hasNext(){return obj.hasNext();}
function init(input){  
  var returnObj = {
    list : input, index : 0, max : input.length,
    hasNext : function(){return (this.index < this.max);},
    next : function(){if(!this.hasNext()){throw "ArrayIndexOutOfBoundsException これ以上ないよ";}else{var returnInput = this.list[this.index];this.index++;return returnInput;}}
  };
  return returnObj;
}
function myout(s){console.log(s);}
function myerr(s){console.error("debug:" + require("util").inspect(s,false,null));}
//[no]要素の扱い。数値型
//不明値、異常時:引数そのまま返す 1:数値へ変換
//2:半角SPで分割 4:半角SPで分割し、数値配列へ
//6:1文字で分割  7:1文字で分割し、数値配列へ
//8:半角SPで結合 9:改行で結合 0:文字なしで結合
function myconv(i,no){try{switch(no){case 1:return parseInt(i);case 2:return i.split(" ");case 4:return i.split(" ").map(Number);case 6:return i.split("");case 7:return i.split("").map(Number);case 8:return i.join(" ");case 9:return i.join("\n");case 0:return i.join("");default:return i;}}catch(e){return i;}}

//↑見なくていいよ！ここまで--------------------------------------------------
function Main(){
	var one = nextIntArray();
	var N = one[0];
	var K = one[1];
	var list = nextIntArray();
	if(false){
		//全部とれる場合
		list.sort(function(a,b){
			return a - b;
		});
		for(var i = 0; i < K - N; i++){
			if(list[0] < 0){
				list.shift();
			}else{
				break;
			}
		}
		var sum = 0;
		for(var i = 0; i < list.length; i++){
			sum += list[i];
		}
		myout(sum);
	}else{
		K = Math.min(K, N);
		//全部とれない場合
		//左端から全部とる、左端からK - 1、右端から1個、・・右端から全部とる
		//↑これをK = 1～できるだけやる？
		//K未満だったら、残回数でマイナスを捨てていく
		var max = 0;
		for(var i = 1; i <= K; i++){
			//i : 取り出す回数(全体)
			
			for(var j = 0; j <= i; j++){//j : 左から取り出す回数
				var maelist = makeClone(list);
				var maeGetlist = [];
				var right = i - j;
				for(var k = 0; k < j; k++){
					maeGetlist.push(maelist.shift());
				}
				for(var k = 0; k < right; k++){
					maeGetlist.push(maelist.pop());
				}
				maeGetlist.sort(function(a,b){
					return a - b;
				});
				var ret = K - i;
				for(var k = 0; k < ret; k++){
					if(maeGetlist[0] < 0){
						maeGetlist.shift();
					}else{
						break;
					}
				}
				var sum = 0;
				for(var k = 0; k < maeGetlist.length; k++){
					sum += maeGetlist[k];
				}
				
				max = Math.max(max, sum);
			}
		}
		myout(max);
	}
}