792C - Divide by Three

C. Divide by Three

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

删掉几个位置,,,使得这个数是3的倍数..删完后长度要最长.

写复杂了,,前导0考虑下..然后直接暴力判断,从后面删前面删各枚举一遍...

#include <bits/stdc++.h>
using namespace std;
const int N=1e6+10;
string str;
string ss;
string sss;
int max1;
string ans;
int vis[10];
int l;
void change() {
if(ss==str)
return;
int l1=ss.size();
if(l1==1) {
if(l1>max1) {
max1=1;
ans=ss;
}
} else {
int x;
int flag=0;
for(int i=0; i<l1; i++)
if(ss[i]!='0') {
flag=1;
x=i;
break;
}
if(flag) {
ss="";
for(int i=x; i<l1; i++)
ss+=sss[i];
l1=ss.size();
}
if(l1>max1) {
max1=l1;
ans=ss;
}
}
}
void work(int m,int k,int kk,int kkk,int kkkk,int kkkkk,int kkkkkk) {
int op=k*1+kk*2+kkk*7+kkkk*4+kkkkk*5+kkkkkk*8;
if(vis[1]>=k&&vis[2]>=kk&&vis[7]>=kkk&&vis[4]>=kkkk&&vis[5]>=kkkkk&&vis[8]>=kkkkkk&&(op%3==m)) {
int qq=0;
int qqq=0;
int qqqq=0;
int qqqqq=0;
int qqqqqq=0;
int qqqqqqq=0;
ss=sss="";
for(int i=0; i<l; i++) {
if(str[i]=='1'&&qq<k) {
qq++;
continue;
}
if(str[i]=='2'&&qqq<kk) {
qqq++;
continue;
}
if(str[i]=='7'&&qqqq<kkk) {
qqqq++;
continue;
}
if(str[i]=='4'&&qqqqq<kkkk) {
qqqqq++;
continue;
}
if(str[i]=='5'&&qqqqqq<kkkkk) {
qqqqqq++;
continue;
}
if(str[i]=='8'&&qqqqqqq<kkkkkk) {
qqqqqqq++;
continue;
}
ss+=str[i];
sss+=str[i];
}
change();
string a;
ss=sss="";
qq=qqq=qqqq=qqqqq=qqqqqq=qqqqqqq=0;
for(int i=l-1; i>=0; i--) {
if(str[i]=='1'&&qq<k) {
qq++;
continue;
}
if(str[i]=='2'&&qqq<kk) {
qqq++;
continue;
}
if(str[i]=='7'&&qqqq<kkk) {
qqqq++;
continue;
}
if(str[i]=='4'&&qqqqq<kkkk) {
qqqqq++;
continue;
}
if(str[i]=='5'&&qqqqqq<kkkkk) {
qqqqqq++;
continue;
}
if(str[i]=='8'&&qqqqqqq<kkkkkk) {
qqqqqqq++;
continue;
}
a+=str[i];
}
int op=a.size();
for(int i=op-1; i>=0; i--) {
ss+=a[i];
sss+=a[i];
}
change();
}
}
int main() {
while(cin>>str) {
l=str.size();
int sum=0;
memset(vis,0,sizeof(vis));
for(int i=0; i<l; i++) {
sum+=str[i]-'0';
vis[str[i]-'0']++;
}
if(sum%3==0) {
string qqqq="";
int qer=-1;
int ans1=0;
for(int i=0; i<l; i++)
if(str[i]!='0') {
qer=i;
break;
} else {
ans1++;
}
if(ans1==l) {
cout<<0<<endl;
continue;
}
for(int i=qer; i<l; i++)
qqqq+=str[i];
cout<<qqqq<<endl;
} else {
max1=-9999999;
ans="";
if(l==1) {
cout<<-1<<endl;
} else {
if(sum%3==1) { //1 4 7 22 16 25 115 112 2221 1111 1114 11122 111112 1111111for(int k=0; k<=1; k++) //1
for(int kk=0; kk<=2; kk++) //2
for(int kkk=0; kkk<=1; kkk++) //7
for(int kkkk=0; kkkk<=1; kkkk++) //4
for(int kkkkk=0; kkkkk<=2; kkkkk++) //5
for(int kkkkkk=0; kkkkkk<=2; kkkkkk++) { //8
work(1,k,kk,kkk,kkkk,kkkkk,kkkkkk);
}
} else if(sum%3==2) { // 2 5 8 11for(int k=0; k<=2; k++) //1
for(int kk=0; kk<=1; kk++) //2
for(int kkk=0; kkk<=2; kkk++) //7
for(int kkkk=0; kkkk<=2; kkkk++) //4
for(int kkkkk=0; kkkkk<=1; kkkkk++) //5
for(int kkkkkk=0; kkkkkk<=1; kkkkkk++) { //8
work(2,k,kk,kkk,kkkk,kkkkk,kkkkkk);
}
}
int op1=ans.size();
int res=0;
for(int i=0; i<op1; i++)
if(ans[i]>='0'&&ans[i]<='9')
res+=ans[i]-'0';
if(res%3==0&&ans!="") {
if(res==0)
cout<<0<<endl;
else
cout<<ans<<endl;
} else
cout<<-1<<endl;
}
}
}
return 0;
}