Toda 2

A group of n friends enjoys playing popular video game Toda 2. There is a rating system describing skill level of each player, initially the rating of the i-th friend is r_i.

The friends decided to take part in the championship as a team. But they should have equal ratings to be allowed to compose a single team consisting of all n friends. So the friends are faced with the problem: how to make all their ratings equal.

One way to change ratings is to willingly lose in some matches. Friends can form a party consisting of two to five (but not more than n) friends and play a match in the game. When the party loses, the rating of each of its members decreases by 1. A rating can't become negative, so r_i = 0 doesn't change after losing.

The friends can take part in multiple matches, each time making a party from any subset of friends (but remember about constraints on party size: from 2 to 5 members).

The friends want to make their ratings equal but as high as possible.

Help the friends develop a strategy of losing the matches so that all their ratings become equal and the resulting rating is maximum possible.

Input

The first line contains a single integer n (2 ≤ n ≤ 100) — the number of friends.

The second line contains n non-negative integers r₁, r₂, ..., r_n (0 ≤ r_i ≤ 100), where r_i is the initial rating of the i-th friend.

Output

In the first line, print a single integer R — the final rating of each of the friends.

In the second line, print integer t — the number of matches the friends have to play. Each of the following t lines should contain n characters '0' or '1', where the j-th character of the i-th line is equal to:

• '0', if friend j should not play in match i,
• '1', if friend j should play in match i.

Each line should contain between two and five characters '1', inclusive.

The value t should not exceed 10⁴, it is guaranteed that such solution exists.

Remember that you shouldn't minimize the value t, but you should maximize R. If there are multiple solutions, print any of them.

Examples

Input

Copy

5
4 5 1 7 4

Output

Copy

1
8
01010
00011
01010
10010
00011
11000
00011
11000

Input

Copy

2
1 2

Output

Copy

0
2
11
11

Input

Copy

3
1 1 1

Output

Copy

1
0

题解：数据比较小，每次操作只用两个数。还要确定最终的状态。

#pragma warning(disable:4996)
#include<map>
#include<queue>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

typedef pair<int, int> P;

const int maxn = 105;

P a[maxn];
int n;
int G[maxn*maxn][maxn];

int main()
{
    while (scanf("%d", &n) != EOF) {
        memset(G, 0, sizeof(G));
        for (int i = 1; i <= n; i++) {
            int tp;
            scanf("%d", &tp);
            a[i] = P(tp, i);
        }
        int cnt = 0;
        while (true) {
            sort(a + 1, a + n + 1);
            if (a[n].first == a[1].first) break;
            cnt++;
            int k = 0;
            for (int i = 2; i <= n; i++) if (a[i].first > a[1].first) k++;
            if (2 <= k && k <= 5 && a[n].first - 1 == a[1].first) {      //最终的状态！！！！！
                for (int i = n - k + 1; i <= n; i++) {
                    a[i].first--;
                    G[cnt][a[i].second] = 1;
                }
                break;
            }
            else {
                if (a[n].first) a[n].first--;
                if (a[n - 1].first) a[n - 1].first--;
                G[cnt][a[n].second] = G[cnt][a[n - 1].second] = 1;
            }
        }
        printf("%d
", a[1].first);
        printf("%d
", cnt);
        for (int i = 1; i <= cnt; i++) {
            for (int j = 1; j <= n; j++) {
                printf("%d", G[i][j]);
            }
            printf("
");
        }
    }
    return 0;
}