There is a complete graph of m vertices. Initially, the edges of the complete graph are uncolored. Sunuke did the following for each i (1 ≤ i ≤ n): Select ai vertices from the complete graph and color all edges connecting the selected vertices with color i. None of the sides were painted in multiple colors. Find the minimum value that can be considered as m.

Constraints

* 1 ≤ n ≤ 5
* 2 ≤ ai ≤ 109

Input


n
a1
.. ..
an


Output

Output the minimum value of m on one line.

Examples

Input

2
3
3


Output

5


Input

5
2
3
4
5
6


Output

12
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <climits>
#include <ctime>
#include <queue>
#include <stack>
#include <algorithm>
#include <list>
#include <vector>
#include <set>
#include <map>
#include <iostream>
#include <deque>
#include <complex>
#include <string>
#include <iomanip>
#include <sstream>
#include <bitset>
#include <valarray>
#include <unordered_set>
#include <iterator>
#include <functional>
#include <cassert>
using namespace std;
typedef long long int ll;
typedef unsigned int uint;
typedef unsigned char uchar;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;

#define REP(i,x) for(int i=0;i<(int)(x);i++)
#define REPS(i,x) for(int i=1;i<=(int)(x);i++)
#define RREP(i,x) for(int i=((int)(x)-1);i>=0;i--)
#define RREPS(i,x) for(int i=((int)(x));i>0;i--)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();i++)
#define RFOR(i,c) for(__typeof((c).rbegin())i=(c).rbegin();i!=(c).rend();i++)
#define ALL(container) (container).begin(), (container).end()
#define RALL(container) (container).rbegin(), (container).rend()
#define SZ(container) ((int)container.size())
#define mp(a,b) make_pair(a, b)
#define pb push_back
#define eb emplace_back
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );

template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }
template<class T> ostream& operator<<(ostream &os, const vector<T> &t) {
os<<"["; FOR(it,t) {if(it!=t.begin()) os<<","; os<<*it;} os<<"]"; return os;
}
template<class T> ostream& operator<<(ostream &os, const unordered_set<T> &t) {
os<<"{"; FOR(it,t) {if(it!=t.begin()) os<<","; os<<*it;} os<<"}"; return os;
}
template<class S, class T> ostream& operator<<(ostream &os, const pair<S,T> &t) { return os<<"("<<t.first<<","<<t.second<<")";}
template<class S, class T> pair<S,T> operator+(const pair<S,T> &s, const pair<S,T> &t){ return pair<S,T>(s.first+t.first, s.second+t.second);}
template<class S, class T> pair<S,T> operator-(const pair<S,T> &s, const pair<S,T> &t){ return pair<S,T>(s.first-t.first, s.second-t.second);}

const int INF = 1<<28;
const double EPS = 1e-8;
const int MOD = 1000000007;


int T, n, m;
vi d;

int dfs(int i, vi v){
	if(i == n) return count_if(ALL(v), [](int i){return i;});
	int res = 15;
	std::function<void(vi&, int, int)> rec = [&](vi &t, int l, int sum){
		if(l >= 15) return;
		if(t.size() == d[i]){
			auto v2 = v;
			for(int j : t) v2[j] |= 1<<i;
			res = min(res, dfs(i+1, v2));
		}else{
			if(v[l]) rec(t, l+1, sum);
			if(!(sum & v[l])){
				t.pb(l);
				rec(t, l+1, sum | v[l]);
				t.pop_back();
			}
		}
	};
	vi t;
	rec(t, 0, 0);
	return res;
}

int main(int argc, char *argv[]){
	ios::sync_with_stdio(false);
	cin >> n;
	d.resize(n);
	REP(i, n) cin >> d[i];
	sort(RALL(d));
	ll ans = 0;
	REP(i, n){
		if(d[i] >= 5){
			ans += d[i] - 5;
			d[i] = 5;
		}
	}
	cout << ans + dfs(0, vi(15)) << endl;
	return 0;
}