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 Python (3.4.3) (judge verdict: Runtime Error)

#include <vector>
#include <math.h>
#include <iostream>
#include <string>
#include <algorithm>
#include <queue>
#include <iomanip>
#include <deque>
#include <cctype>
#include <unordered_map>

using namespace std;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int,int>;
using pll = pair<long, long>;
template <class T> using V = vector<T>;
template <class T> using VV = V<V<T>>;
#define rep(i, n) for(int i = 0; i<n; i++)
#define rep_s(i, start, n) for(int i= start;  i<n; i++)
#define rep_down(i, n) for(int i = n-1; i>=0; i--)
#define rep_down_s(i, n, start) for(int i = n-1; i>=start; i--)
#define ALL(a) (a).begin(), (a).end()
#define ALL_str(a) (a).cbegin(), (a).cend()

string alphabet_upper = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string alphabet_lower  = "abcdefghijklmnopqrstuvwxyz";
string numbers = "0123456789";
/*
for (iterator = hash.begin(); iterator != hash.end(); iterator++)
{
// ここで、要素毎の操作を行います。
std::pair<std::string, CMyClass> element = *iterator;

}
*/
/*
using edge = struct {int to; ll cost;};
vector<edge> tree[limit];
ll depth[limit];

void dfs(int v, int p, ll d){
  depth[v] = d;
  //cout <<"v: " << v << endl;
  for (auto &e: tree[v]){
    //cout << "e.to: " << e.to << endl;
    if (e.to == p) continue;
    dfs(e.to, v, d+e.cost);
  }
}
*/
int keta_sum(int num){
  int dig, sum = 0;
  while(num){
    dig = num % 10;
    sum = sum + dig;
    num = num / 10;
  }
  return sum;
}

ll binary_search(ll left_key, ll border, ll low, ll high) {
     // 配列 a の左端と右端
    ll mid = (low+high)/2;
    while (high > low+1) {
       mid = (low+high)/2;
        if (left_key*mid>border) high = mid;
        else if (left_key*mid<border) low = mid;
        else {
          mid =mid-1;
          break;
        }
    }
    return mid;
}

bool IsPrime(ll a){
  ll b = sqrt(a);
  b++;
  while(b-->=2){
    if (a%b==0) return false;
  }
  return true;
}

ll gcd(ll a, ll b){
  ll tmp;
  if(a<b){
    tmp = a;
    a = b;
    b = tmp;
  }

  ll r = a % b;
  while(r!=0){
    a = b;
    b = r;
    r = a % b;
  }
  return b;
}

struct UnionFind {
    vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

    UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
        for(int i = 0; i < N; i++) par[i] = i;
    }

    int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y) { // xとyの木を併合
        int rx = root(x); //xの根をrx
        int ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
    }

    bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
};

int main(void){
  ll N, M, A, B, C, D, E, F, K,L, W, H, P, Q, R, X, Y;
  ll ans;
  string S, T;
  cin >> S;
  ll count0 = 0, count1 = 0;
  rep(i, S.length()){
    if (S[i] == '0') count0++;
    else count1++;
  }
  cout << min(count0, count1)*2 << endl;

}