Divide by Three

A positive integer number n is written on a blackboard. It consists of not more than 10⁵ digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.

The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.

Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.

If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.

Input

The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10¹⁰⁰⁰⁰⁰).

Output

Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print  - 1.

Example

Input

1033

Output

33

Input

10

Output

0

Input

11

Output

-1

Note

In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two.

题意：一个没有前导0的数字中，删若干个数字使它能够被3整除，要求最长的输出

思路：

第一步：求所有数字的和模3，取余数

第二步：从后往前每个数字查询，分为两种，删去余数与第一步余数相同的数字，删去两个余数与第一步不同的数字（不考虑0，1的时候删2,2的时候删1）。

由于从后往前查询，所以得到的一定是最优解，如果删去数字时产生了前导0，即删去了的一位，则证明第一位必须要删去，否则无法得到被3整除的数字。

#include
#include
#include
#include
#include
#define maxn 100010
using namespace std;

int n[maxn];
char str[maxn];
char ans1[maxn];//第一种方案
char ans2[maxn];//第二种方案

int main()
{
    gets(str);
    int len=strlen(str);
    int p1=0,p2=0,p3=0,sum=0,mod; //p1表示余数相同的第一个数字 ， p2,p3表示余数不同的两个数字
    for(int i=1;i<=len;i++){
        n[i]=str[i-1]-48;
        sum+=n[i];
    }
    mod=sum%3;
    if(mod==0){   //已经是3的倍数的情况
        printf("%s",str);
        return 0;
    }
    for(int i=len;i>=1;i--){
        if(n[i]%3==mod){
            p1=i;
            break;
        }
    }
    for(int i=len;i>=1;i--){
        if(n[i]%3==3-mod){
            p3=i;
            for(int j=i-1;j>=1;j--){
                if(n[j]%3==3-mod){
                    p2=j;
                    break;
                }
            }
            break;
        }
    }
    if(p1!=0){ 
        for(int i=1;i=len2)
        for(int i=0;i