Jeff's got n cards, each card contains either digit 0, or digit 5. Jeff can choose several cards and put them in a line so that he gets some number. What is the largest possible number divisible by 90 Jeff can make from the cards he's got?

Jeff must make the number without leading zero. At that, we assume that number 0 doesn't contain any leading zeroes. Jeff doesn't have to use all the cards.

Input

The first line contains integer n (1 ≤ n ≤ 10³). The next line contains n integers a₁, a₂, ..., a_n (a_i = 0 or a_i = 5). Number a_i represents the digit that is written on the i-th card.

Output

In a single line print the answer to the problem — the maximum number, divisible by 90. If you can't make any divisible by 90 number from the cards, print -1.

Examples Input Copy

4
5 0 5 0

Output Copy

0

Input Copy

11
5 5 5 5 5 5 5 5 0 5 5

Output Copy

5555555550

Note

In the first test you can make only one number that is a multiple of 90 — 0.

In the second test you can make number 5555555550, it is a multiple of 90.

 

%9的性質還記得嗎？個位數字相加看%9==0即可；

注意題目是90；簡單模擬一下即可；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
    ll x = 0;
    char c = getchar();
    bool f = false;
    while (!isdigit(c)) {
        if (c == '-') f = true;
        c = getchar();
    }
    while (isdigit(c)) {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return f ? -x : x;
}

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
    if (!b) {
        x = 1; y = 0; return a;
    }
    ans = exgcd(b, a%b, x, y);
    ll t = x; x = y; y = t - a / b * y;
    return ans;
}
*/

int a[maxn];
bool cmp(int a, int b) { return a > b; }
int main() {
    //ios::sync_with_stdio(0);
    int n; int cnt = 0; cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i]; if (a[i] == 0)cnt++;
    }
    sort(a + 1, a + 1 + n, cmp);
    int sum = 0; int tot = 0; bool fg = 0; int maxx = 0;
    for (int i = 1; i <= n; i++) {
        if (a[i] == 0)break;
        sum += a[i]; tot++;
    //	cout << sum << endl;
        if (sum % 9 == 0) {
            fg = 1; maxx = max(maxx, tot);
        }
            
    }
    
    if (!fg&&cnt==0)cout << -1 << endl;
    else {
        if (maxx == 0&&cnt) {
            cout << 0 << endl; return 0;
        }
        if (cnt == 0) {
            cout << -1 << endl; return 0;
        }
        for (int i = 1; i <= maxx; i++)cout << 5;
        for (int i = 1; i <= cnt; i++)cout << 0;
    }

    return 0;
}