Educational Codeforces Round 18

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.

Examples

input

output

input

output

input

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：

最大长度是1e5的数字，问去掉最少的数之后，他能被3整除，输出他。

代码：

//模拟。所有位数之和sum，sum%3只可能是0,1,2，是0就直接去前导0输出，
//不是0就从后向前看有没有s[i]%3==sum%3，有就去掉s[i]，把新的s存入ch1,
//再从后向前看有没有(s[i]+s[j])==sum%3,有就去掉s[i],s[j]，把新的s存入ch2,
//最后去掉ch1,ch2的前导0，输出位数多的那个。
#include<bits/stdc++.h>
using namespace std;
char s[100010],ch[100010],ch1[100010],ch2[100010];
int main()
{
    scanf("%s",s);
    int len=strlen(s),sum=0,flag=0;
    for(int i=0;i<len;i++){
        sum+=s[i]-'0';
        if(s[i]=='0') flag=1;
    }
    int tmp=sum%3;
    if(tmp==0){
        int nu=0,j=0;
        while(nu<len-1&&s[nu]=='0'){
            nu++;
        }
        for(int i=nu;i<=len-1;i++) ch[j++]=s[i];
        printf("%s
",ch);
    }
    else {
        int p1=-1,p2=-1,p3=-1;
        for(int i=len-1;i>=0;i--){
            if((s[i]-'0')%3==tmp){
                p1=i;break;
            }
        }
        for(int i=len-1;i>=1;i--){
            if((s[i]-'0')%3==0) continue;
            for(int j=i-1;j>=0;j--){
                int x=s[i]-'0'+s[j]-'0';
                if(x%3==tmp){
                    p2=i;p3=j;break;
                }
            }
            if(p2!=-1&&p3!=-1) break;
        }
        int x1=0,x2=0;
        if(p1!=-1){
            int j=0,nu=0;
            for(int i=0;i<=len-1;i++){
                if(i==p1)continue;
                ch[j++]=s[i];
            }
            while(nu<j-1&&ch[nu]=='0'){
                nu++;
            }
            for(int i=nu;i<=j-1;i++)
                ch1[x1++]=ch[i];
        }
        if(p2!=-1&&p3!=-1){
            int j=0,nu=0;
            for(int i=0;i<=len-1;i++){
                if(i==p2||i==p3) continue;
                ch[j++]=s[i];
            }
            while(nu<j-1&&ch[nu]=='0'){
                nu++;
            }
            for(int i=nu;i<=j-1;i++)
                ch2[x2++]=ch[i];
        }
        if(!flag&&x1==0&&x2==0) printf("-1
");
        else if(x1==0&&x2==0) printf("0
");
        else if(x1>=x2) printf("%s
",ch1);
        else printf("%s
",ch2);
    }
    return 0;
}