Problem:
Score : 300 points

Problem Statement

There are N cubes stacked vertically on a desk.

You are given a string S of length N. The color of the i-th cube from the bottom is red if the i-th character in S is 0, and blue if that character is 1.

You can perform the following operation any number of times: choose a red cube and a blue cube that are adjacent, and remove them. Here, the cubes that were stacked on the removed cubes will fall down onto the object below them.

At most how many cubes can be removed?

Constraints

1 \leq N \leq 10^5

|S| = N

Each character in S is 0 or 1.

Input

Input is given from Standard Input in the following format:

S

Output

Print the maximum number of cubes that can be removed.

Sample Input 1

0011

Sample Output 1

4

All four cubes can be removed, by performing the operation as follows:

Remove the second and third cubes from the bottom. Then, the fourth cube drops onto the first cube.

Remove the first and second cubes from the bottom.

Sample Input 2

11011010001011

Sample Output 2

12

Sample Input 3

0

Sample Output 3

0

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

/* ex 9_2
   person_A */

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define SIZE 100001

//リストのノード（スタック内の各データ）の定義
typedef struct node{
    char elem;
    struct node *next;
} node;

//スタックにデータを追加
void push(node  **first, char elem){
    //追加する分の領域を確保
    node *new = calloc(1, sizeof(node));
    if(new == NULL) exit(1);
    new->elem = elem;
    new->next = *first;
    *first = new;
}

//スタックからデータを取り出す
char pop(node **first){
    char elem;
    node *tmp = *first;
    if(tmp == NULL) exit(1);
    elem = tmp->elem;
    *first = tmp->next;
    //取り出したノードのメモリを解放
    free(tmp);
    return elem;
}


int main(void){
    //最初のノードの指す先
    node *first = NULL;
    char cubes[SIZE],top;
    int i,count=0,strlength;
    //赤色のキューブ'0'と青色のキューブ'1'を表す文字列の入力
    scanf("%s",cubes);
    strlength=strlen(cubes);

    //下の方からキューブをスタックに入れていく
    for(i=0;i<strlength;i++){
        //机の上にキューブがなければ、一つを置くだけ
        if(first==NULL){
            push(&first,cubes[i]);
        }
        else{
            top=pop(&first);
            //先頭と異なる色であれば、カウントする
            if(top!=cubes[i]){
                count+=2;
            }
            //先頭と同じ色であれば、二つを机の上に置き戻す
            else{
                push(&first,top);
                push(&first,cubes[i]);
            }
        }
    }
    
    //結果の出力
    printf("%d\n",count);

    return 0;
}