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)

/* ex8_2 
melo */
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define SIZE 1000001
typedef struct node{
    char elem;//文字
    struct node *next;//次のノードへのパス
} node;
void push(node **first,char elem);//スタックにelemを格納
char pop(node **first);//スタックの一番上から取り出す

int main(void){
    int i,count=0;//for文用,キューブの数
    char color[SIZE];//キューブの色の配列
    node *first=NULL;//最初のノードのさす先
    scanf("%s",color);//
    for(i=0;i<strlen(color);i++){//
        if(first==NULL){//スタックに何も入っていないとき
            push(&first,color[i]);//スタックにcolor[i]を格納
        }
        else if(color[i]==first->elem){//真上の色と一緒だった場合
            push(&first,color[i]);//スタックにcolor[i]を格納
        }
        else{//色が違った場合
            pop(&first);//取り除く
            count+=2;//キューブを二つ取り除く
        }
    }
    printf("%d\n",count);//何個キューブを取り除いたか
    return 0;//正常終了
}

void push(node **first,char elem){
    node *new =calloc(1,sizeof(node));//追加する分の領域を確保
    if(new==NULL)exit(1);//確保できなかった場合
    new->elem=elem; //要素を一番上に追加
    new->next=*first;
    *first=new;
}
char pop(node **first){
    char elem;//取り出す要素
    node *tmp =*first;//一番上
    if(tmp==NULL)exit(1);//何も入っていなかったとき
    elem=tmp->elem;//
    *first=tmp->next;//一番上を一つ下げる
    free(tmp);//取り出したノードのメモリを解放
    return elem;//取り出した要素を返す
}