
#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES
#include <cfloat>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <iostream>
#include <memory>
#include <string>

#include <algorithm>
#include <complex>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <bitset>

using namespace std;

#ifdef _MSC_VER
#define __typeof__ decltype
#define strtok_r strtok_s
#endif

#define ITER(c)             __typeof__((c).begin())
#define FOREACH(it, c)      for (ITER(c) it=(c).begin(); it != (c).end(); ++it)
#define RITER(c)            __typeof__((c).rbegin())
#define RFOREACH(it, c)     for (RITER(c) it=(c).rbegin(); it != (c).rend(); ++it)
#define REP(i, n)           REPEAT(i, 0, n)
#define RREP(i, n)          RREPEAT(i, 0, n)
#define REPEAT(i, k, n)     for(int i = (k); i < ((k)+(n)); ++i)
#define RREPEAT(i, k, n)    for(int i = (k)+(n)-1; i >= (k); --i)
#define FROMTO(i,f,t)       for(int i = f; i < t; i++)
#define ALL(c)              (c).begin(), (c).end()
#define LLPOW(p1,p2)        ((ll)pow((double)(p1), (int)p2))
#define ESIZEOF(A)          (sizeof(A)/sizeof((A)[0]))
#define CIN_NO_SYNC         do { cin.tie(0); ios_base::sync_with_stdio(false); } while(0)
#define GETSTR(p)           fgets((p), sizeof(p), stdin)
#define CHOMP(p)            do{ char *_q = (p) + strlen(p)-1; if(*_q == '\n' || *_q == '\r') *_q = 0; } while(0)
#define FILL(m,v)           memset(m, v, sizeof(m))
#define mp                  make_pair
#define pb                  push_back
template<class _T> _T MAX(_T p1,_T p2){return (p1>p2)?p1:p2;} 
template<class _T> _T MIN(_T p1,_T p2){return (p1<p2)?p1:p2;} 
template <class _Tv> inline string join(const _Tv &v,string sep=" "){
    ostringstream s;FOREACH(it,v){if(it!=v.begin())s<<sep;s<<*it;};return s.str();
}

typedef long long ll;
typedef unsigned long long ull;
typedef double D;
typedef complex<D> P;
typedef vector<int> VI;
typedef vector<VI> VVI;
#define X real()
#define Y imag()
#define EPS (1e-9)
#define DEQ(p1,p2) (abs((p1)-(p2)) < EPS)

#ifdef _DEBUG
template<class _Tv> inline void _prfe(const char *_n,const _Tv _c,bool _p=false){
    ITER(_c) _it=_c.begin();
    if(_p){cout<<_n<<" = ["<<endl;for(;_it!=_c.end();++_it)cout<<"    "<<*_it<<endl;cout<<"]"<<endl; }
    else{cout<<_n<<" = [ "<<*_it++;for(;_it!=_c.end();++_it)cout<<", "<<*_it;cout<<" ]"<<endl; }
}
ostream &operator<<(ostream &os, const pair<int,int> &p){return(os<<"("<<p.first<<","<<p.second<<")");}
#define pf printf
#define pr(n) do{cout<<#n" = "<<(n)<<endl;}while(0)
#define prfe(n) _prfe(#n, n)
#define prfep(n) _prfe(#n, n, true)
#define dbgchk(n) do{if(n)throw;}while(0)
#else
#define pf(...) /* */
#define pr(...) /* */
#define prfe(...) /* */
#define prfep(...) /* */
#define dbgchk(...) /* */
#endif

int solve(VI v){
    int t = 0, ac = 0, max = 0;
    FOREACH(it, v){
        if(*it > 10) *it = 10;
        if(*it == 1) ac++;
        t += *it;
    }
    max = MAX(max, (t>21?0:t));
    REP(i,ac){
        t += 10;
        max = MAX(max, (t>21?0:t));
    }
    return max;
}

int main(){
    string s;
    while(getline(cin,s)){
        stringstream ss(s);
        int n;
        ss >> n;
        if(n == 0) break;
        VI v;
        v.pb(n);
        while(ss >> n) v.pb(n);
        cout << solve(v) << endl;
    }
    return 0;
}
### Prompt

Please create a solution in CPP to the following problem:

Blackjack is a type of card game played in casinos, where the game is played using cards with numbers from 1 to 13. The score of each card is decided as follows.

* 1 is 1 point or 11 points
* From 2 to 9, the score is as written.
* 10 points from 10 to 13



There are several participants in this game, including parents, each with several sets of cards. This set of cards is called a hand. The hand score is the total of the card scores. The calculation shall be performed as follows.

* If the total score of the cards is greater than 21, the score of the hand is set to 0.
* As the score of the card, 1 may be calculated as 1 point or 11 points, but the one with the maximum hand score shall be selected.



Create a program that uses the information of the cards dealt as input and outputs the score of the hand.



Input

A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:


c1 c2 ... cn


The integer ci (1 ≤ ci ≤ 13) written on the i-th card is given to each line, separated by blanks. The number of cards n does not exceed 100.

The number of datasets does not exceed 200.

Output

The score of the hand is output to one line for each data set.

Example

Input

1
7 7 7
7 7 8
12 1
10 1 1
0


Output

11
21
0
21
12

### Response

```cpp

#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES
#include <cfloat>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <iostream>
#include <memory>
#include <string>

#include <algorithm>
#include <complex>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <bitset>

using namespace std;

#ifdef _MSC_VER
#define __typeof__ decltype
#define strtok_r strtok_s
#endif

#define ITER(c)             __typeof__((c).begin())
#define FOREACH(it, c)      for (ITER(c) it=(c).begin(); it != (c).end(); ++it)
#define RITER(c)            __typeof__((c).rbegin())
#define RFOREACH(it, c)     for (RITER(c) it=(c).rbegin(); it != (c).rend(); ++it)
#define REP(i, n)           REPEAT(i, 0, n)
#define RREP(i, n)          RREPEAT(i, 0, n)
#define REPEAT(i, k, n)     for(int i = (k); i < ((k)+(n)); ++i)
#define RREPEAT(i, k, n)    for(int i = (k)+(n)-1; i >= (k); --i)
#define FROMTO(i,f,t)       for(int i = f; i < t; i++)
#define ALL(c)              (c).begin(), (c).end()
#define LLPOW(p1,p2)        ((ll)pow((double)(p1), (int)p2))
#define ESIZEOF(A)          (sizeof(A)/sizeof((A)[0]))
#define CIN_NO_SYNC         do { cin.tie(0); ios_base::sync_with_stdio(false); } while(0)
#define GETSTR(p)           fgets((p), sizeof(p), stdin)
#define CHOMP(p)            do{ char *_q = (p) + strlen(p)-1; if(*_q == '\n' || *_q == '\r') *_q = 0; } while(0)
#define FILL(m,v)           memset(m, v, sizeof(m))
#define mp                  make_pair
#define pb                  push_back
template<class _T> _T MAX(_T p1,_T p2){return (p1>p2)?p1:p2;} 
template<class _T> _T MIN(_T p1,_T p2){return (p1<p2)?p1:p2;} 
template <class _Tv> inline string join(const _Tv &v,string sep=" "){
    ostringstream s;FOREACH(it,v){if(it!=v.begin())s<<sep;s<<*it;};return s.str();
}

typedef long long ll;
typedef unsigned long long ull;
typedef double D;
typedef complex<D> P;
typedef vector<int> VI;
typedef vector<VI> VVI;
#define X real()
#define Y imag()
#define EPS (1e-9)
#define DEQ(p1,p2) (abs((p1)-(p2)) < EPS)

#ifdef _DEBUG
template<class _Tv> inline void _prfe(const char *_n,const _Tv _c,bool _p=false){
    ITER(_c) _it=_c.begin();
    if(_p){cout<<_n<<" = ["<<endl;for(;_it!=_c.end();++_it)cout<<"    "<<*_it<<endl;cout<<"]"<<endl; }
    else{cout<<_n<<" = [ "<<*_it++;for(;_it!=_c.end();++_it)cout<<", "<<*_it;cout<<" ]"<<endl; }
}
ostream &operator<<(ostream &os, const pair<int,int> &p){return(os<<"("<<p.first<<","<<p.second<<")");}
#define pf printf
#define pr(n) do{cout<<#n" = "<<(n)<<endl;}while(0)
#define prfe(n) _prfe(#n, n)
#define prfep(n) _prfe(#n, n, true)
#define dbgchk(n) do{if(n)throw;}while(0)
#else
#define pf(...) /* */
#define pr(...) /* */
#define prfe(...) /* */
#define prfep(...) /* */
#define dbgchk(...) /* */
#endif

int solve(VI v){
    int t = 0, ac = 0, max = 0;
    FOREACH(it, v){
        if(*it > 10) *it = 10;
        if(*it == 1) ac++;
        t += *it;
    }
    max = MAX(max, (t>21?0:t));
    REP(i,ac){
        t += 10;
        max = MAX(max, (t>21?0:t));
    }
    return max;
}

int main(){
    string s;
    while(getline(cin,s)){
        stringstream ss(s);
        int n;
        ss >> n;
        if(n == 0) break;
        VI v;
        v.pb(n);
        while(ss >> n) v.pb(n);
        cout << solve(v) << endl;
    }
    return 0;
}
```