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: Runtime Error)

/*  ex9_2
    Y_Aizaki  */

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

#define SIZE 100000

typedef struct node{
    char elem;
    struct node *next;
}stack;

void push(stack **first, char elem);
char pop(stack **first);

int main(void){
    stack *stk;
    char input[SIZE];
    char *input_p = input;
    int removed = 0; // 除かれたキューブの数

    scanf("%s", input);
    stk = NULL; // stkの初期化

    while (*input_p != '\0'){ // inputを前から見ていくループ
        if (stk == NULL){ // stkが空の時
            push(&stk, *input_p);
            printf("ue input_p:%p elem:%c next:%p\n", input_p, stk->elem, stk->next);
        }
        else{ // stkに要素があるとき
            if (stk->elem != *input_p){ // inputで見ている要素と、stkの中身が違う時
                pop(&stk); // stkから消去
                removed += 2;
                printf("na input_p:%p elem:%c next:%p\n", input_p, stk->elem, stk->next);
            }
            else{ // inputで見ている要素と、stkの中身が同じ時
                push(&stk, *input_p); // stkに追加
                printf("sh input_p:%p elem:%c next:%p\n", input_p, stk->elem, stk->next);
            }
        }
        input_p++;
    }

    printf("%d\n", removed);

    return 0;
}

void push(stack **first, char elem){
    stack *new = calloc(1, sizeof(stack));
    if (new == NULL) exit(1);
    new->elem = elem;
    new->next = *first;
    *first = new;
}


char pop(stack **first){
    char elem;
    stack *tmp = *first;
    if (tmp == NULL) exit(1);
    elem = tmp->elem;
    *first = tmp->next;

    free(tmp);
    return elem;
}