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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1ll;while(b){if(b&1)ans=ans*a%MOD;a=a*a%MOD;b>>=1;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
string x;
int main()
{
	//    freopen("C:\\Users\\DH_M\\Desktop\\code_io\\in.txt.txt","r",stdin);
	//    freopen("C:\\Users\\DH_M\\Desktop\\code_io\\out.txt.txt","w",stdout);
	cin>>x;
	int cnt=0;
	while(1)
	{
		int id=x.find("01");
		if(id==-1)
		{
			id=x.find("10");
			if(id==-1)
			{
				break;
			}
		}
		x.erase(id,2);
		cnt+=2;
	}
	cout<<cnt<<endl;
	
	
    return 0;
}

inline void getInt(int* p) {char ch;do {ch = getchar();} 
while (ch == ' ' || ch == '\n');if (ch == '-') {*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {*p = *p * 10 - ch + '0';}}
else {*p = ch - '0';while ((ch = getchar()) >= '0' && ch <= '9') 
{*p = *p * 10 + ch - '0';}}}