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)

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


#define SIZE 1000001

typedef struct node{
    char num;
    struct node *next;//後の要素へのポインタ
}stack;



void push(stack **first,  char number){
    stack *new = calloc(1,sizeof(stack));//メモリ確保
    if(new == NULL){
        exit(1);
    }
    new->num = number;//newの指すアドレスのnumの要素にnumber代入
    new->next = *first;//新しく作ったノードの次のノードをfirstが指すノードにする。
    *first = new;//一番手前のノードをfirstとして更新
    
}

char pop(stack **first){
    char elem;
    stack *tmp = *first;//保存場所用意。

    if(tmp == NULL){//メモリ確保
        exit(1);
    }

    elem = tmp->num;
    *first = tmp->next;//firstの値を、今指すノード(先頭だったもの)の次のノードに変える。
    free(tmp);//メモリ開放

    return elem;
    
}


int main(void){
    char sentence[SIZE];
    int i;
    int collected = 0;
    stack *stk;

    stk = NULL;//初めはNULLを指すようにする。

    scanf("%s", sentence);
    for(i=0; sentence[i] != '\0'; i++){
        if(stk == NULL){//1文字目だった時の処理
            push(&stk, sentence[i]);
        }
        else{
            if(sentence[i] != stk->num){//異なる色だった時⇒一回取り出し、収集数+2
                pop(&stk);
                collected+=2;
            }
            else{//同じ色だった時⇒その要素を入れる
                push(&stk, sentence[i]);
            }
        }
    }      
    
    //合計出力
    printf("%d\n", collected);

    return 0;
}