Divide by Three CodeForces - 792C

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

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.

给定一个串，要求该数去掉尽可能少的位，使得剩下的数可以被3整除

就是所有位的和都可以被3整除

其实最多减两个 如果 sum%3==1 那么可以减去两个 %3==2的 和一个 %3=1的

反之亦然

 

#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <set>
#include <iostream>
#include <map>
#include <stack>
#include <string>
#include <vector>
#define  pi acos(-1.0)
#define  eps 1e-6
#define  fi first
#define  se second
#define  lson l,m,rt<<1
#define  rson m+1,r,rt<<1|1
#define  bug         printf("******\n")
#define  mem(a,b)    memset(a,b,sizeof(a))
#define  fuck(x)     cout<<"["<<"x="<<x<<"]"<<endl
#define  f(a)        a*a
#define  sf(n)       scanf("%d", &n)
#define  sff(a,b)    scanf("%d %d", &a, &b)
#define  sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define  sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
#define  FIN         freopen("DATA.txt","r",stdin)
#define  gcd(a,b)    __gcd(a,b)
#define  lowbit(x)   x&-x
#pragma  comment (linker,"/STACK:102400000,102400000")
using namespace std;
typedef long long  LL;
typedef unsigned long long ULL;
const int INF = 0x7fffffff;
const LL  INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int mod = 1e9 + 7;
const int maxn = 1e4 + 10;
string s1, s2, s3;
int cal(string s) {
    int sum = 0;
    for (int i = 0 ; i < s.size() ; i++) sum = (sum + s[i] - '0') % 3;
    if (!s.size()) return 1;
    return sum;
}
void del(string &s) {
    while(s[0] == '0' && s.size() > 1) s.erase(0, 1);
}
int main() {
    cin >> s1;
    int tot = cal(s1), num1 = 1, num2 = 2, p = 3 - tot;
    s2 = s3 = s1;
    if (!tot) {
        cout << s1 << endl;
        return 0;
    }
    for (int i = s1.size() ; i >=0 ; i--) {
        if ((s2[i] - '0')%3 == tot && num1 ) {
            s2.erase(i, 1);
            num1--;
        }
        if ((s3[i] - '0')%3 == p  && num2) {
            s3.erase(i, 1);
            num2--;
        }
    }
    num1 = cal(s2), num2 = cal(s3);
    del(s2), del(s3);
    if (num1 && num2) return 0 * printf("-1\n");
    if (!num1 && !num2) {
        if (s2.size() > s3.size()) cout << s2 << endl;
        else cout << s3 << endl;
    } else if (!num1) cout << s2 << endl;
    else cout << s3 << endl;
    return 0;
}