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
   Sagzar1993 */

#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 cnt = 0; //除去可能なキューブの数
    char cube; //キューブの色を表す文字
    //最初のノードの指す先
    node *cubeFirst = NULL;
    
    //１文字ずつ入力してスタックを操作
    while(1){
        scanf("%c", &cube);
        //文字列の最後に来たら終了
        if(cube == '\n'){
            break;
        }
        if(cubeFirst == NULL){
            push(&cubeFirst, cube); //１文字目をスタックに追加
        }else if((cubeFirst->elem == '0' && cube == '1') || (cubeFirst->elem == '1' && cube == '0')){
            pop(&cubeFirst); //入力されたcubeをスタックの先頭と比較し、０と１の組ならば除去可能
            cnt += 2; //取り出したスタックの先頭および入力の２文字が除去可能なので、cntを＋２
        }else{
            push(&cubeFirst, cube); //０同士、１同士ならば除去不可なのでそのままスタックに追加
        }
    }
    
    printf("%d\n", cnt); //除去可能なキューブ数cntを出力
    return 0;
}