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)

/*ex9-2
  Sweet_Sweet_Soul*/


#include <stdio.h>
#include <stdlib.h>

//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){
    int count=0;/*答えのカウント*/
    char cube[100001];/*100000文字なのでnull文字含めて確保*/
    char *ptr;/*文字列操作用のポインタ*/
    char tmp;/*pop用の変数*/
    node *list=NULL;/*リスト用のポインタ*/
    scanf("%s",cube);
    ptr=cube;
    while(*ptr!='\0'){
        if(list==NULL || list->elem==*ptr){ /*listの一番上とptrの現在地が一緒 or listが空の場合*/
            push(&list,*ptr);
        }
        else{ /*listの一番上とptrの現在地が一緒*/
            pop(&list);/*listとcubeで一つづつ除去*/
            count=count+2;/*取り出した数をカウント*/
        }
        ptr++;
    }
    printf("%d",count);
    return 0;
}