Problem: Ivan is reading a book about tournaments. He knows that a tournament is an oriented graph with exactly one oriented edge between each pair of vertices. The score of a vertex is the number of edges going outside this vertex. 

Yesterday Ivan learned Landau's criterion: there is tournament with scores d1 ≤ d2 ≤ ... ≤ dn if and only if <image> for all 1 ≤ k < n and <image>.

Now, Ivan wanna solve following problem: given a set of numbers S = {a1, a2, ..., am}, is there a tournament with given set of scores? I.e. is there tournament with sequence of scores d1, d2, ..., dn such that if we remove duplicates in scores, we obtain the required set {a1, a2, ..., am}? 

Find a tournament with minimum possible number of vertices. 

Input

The first line contains a single integer m (1 ≤ m ≤ 31).

The next line contains m distinct integers a1, a2, ..., am (0 ≤ ai ≤ 30) — elements of the set S. It is guaranteed that all elements of the set are distinct.

Output

If there are no such tournaments, print string "=(" (without quotes).

Otherwise, print an integer n — the number of vertices in the tournament.

Then print n lines with n characters — matrix of the tournament. The j-th element in the i-th row should be 1 if the edge between the i-th and the j-th vertices is oriented towards the j-th vertex, and 0 otherwise. The main diagonal should contain only zeros.

Examples

Input

2
1 2


Output

4
0011
1001
0100
0010


Input

2
0 3


Output

6
000111
100011
110001
011001
001101
000000

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
inline void chkmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
inline int readChar();
template <class T = int>
inline T readInt();
template <class T>
inline void writeInt(T x, char end = 0);
inline void writeChar(int x);
inline void writeWord(const char *s);
static const int buf_size = 4096;
inline int getChar() {
  static char buf[buf_size];
  static int len = 0, pos = 0;
  if (pos == len) {
    pos = 0, len = fread(buf, 1, buf_size, stdin);
  }
  if (pos == len) {
    return -1;
  }
  return buf[pos++];
}
inline int readChar() {
  int c = getChar();
  while (c <= 32) {
    c = getChar();
  }
  return c;
}
template <class T>
inline T readInt() {
  int s = 1, c = readChar();
  T x = 0;
  if (c == '-') s = -1, c = getChar();
  while ('0' <= c && c <= '9') x = x * 10 + c - '0', c = getChar();
  return s == 1 ? x : -x;
}
static int write_pos = 0;
static char write_buf[buf_size];
inline void writeChar(int x) {
  if (write_pos == buf_size)
    fwrite(write_buf, 1, buf_size, stdout), write_pos = 0;
  write_buf[write_pos++] = x;
}
template <class T>
inline void writeInt(T x, char end) {
  if (x < 0) writeChar('-'), x = -x;
  char s[24];
  int n = 0;
  while (x || !n) s[n++] = '0' + x % 10, x /= 10;
  while (n--) writeChar(s[n]);
  if (end) writeChar(end);
}
inline void writeWord(const char *s) {
  while (*s) writeChar(*s++);
}
struct Flusher {
  ~Flusher() {
    if (write_pos) fwrite(write_buf, 1, write_pos, stdout), write_pos = 0;
  }
} flusher;
const int MAXN = 1001;
pair<int, int> arr[3001];
char ans[3001][3001];
int A[MAXN];
int n;
bool get228() {
  for (int i = 0; i < n; i++) {
    arr[i].first = A[i];
    arr[i].second = i;
  }
  for (int i = 0; i < n; i++) {
    sort(arr + i, arr + n);
    int j = i + 1;
    while (j < n) {
      if (arr[i].first > 0) {
        arr[i].first -= 1;
        ans[arr[i].second][arr[j].second] = 1;
        ans[arr[j].second][arr[i].second] = 0;
        j++;
      } else {
        if (arr[j].first > 0) {
          arr[j].first -= 1;
          ans[arr[i].second][arr[j].second] = 0;
          ans[arr[j].second][arr[i].second] = 1;
          j++;
        } else {
          break;
        }
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (arr[i].first) {
      assert(false);
      return false;
    }
  }
  return true;
}
int m;
int a[MAXN];
bool dp[32][12000][151];
int pr[32][12000][151];
int main() {
  cin >> m;
  for (int i = 0; i < m; i++) {
    a[i] = readInt();
  }
  sort(a, a + m);
  dp[0][0][0] = true;
  for (int i = 0; i < m; i++) {
    for (int sum = 0; sum < 12000; sum++) {
      for (int cnt = i; cnt <= 150; cnt++) {
        if (dp[i][sum][cnt]) {
          int cursum = sum;
          int curcnt = cnt;
          for (int j = 1; j <= 150; j++) {
            cursum += a[i];
            curcnt++;
            if (cursum < 12000 && cursum >= (curcnt * (curcnt - 1)) / 2) {
              dp[i + 1][cursum][curcnt] = true;
              pr[i + 1][cursum][curcnt] = j;
            } else {
              break;
            }
          }
        }
      }
    }
  }
  vector<int> f;
  bool ff = true;
  for (int cnt = m; cnt <= 150; cnt++) {
    int sum = (cnt * (cnt - 1) / 2);
    if (dp[m][sum][cnt]) {
      int curm = m;
      int cursum = sum;
      int curf = cnt;
      while (curm) {
        int g = pr[curm][cursum][curf];
        for (int i = 0; i < g; i++) {
          f.push_back(a[curm - 1]);
        }
        curm--;
        cursum -= g * a[curm];
        curf -= g;
      }
      ff = false;
      break;
    }
  }
  assert(!ff);
  reverse((f).begin(), (f).end());
  n = (int)(f).size();
  for (int i = 0; i < n; i++) {
    A[i] = f[i];
  }
  if (get228()) {
    writeInt(n, '\n');
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        writeInt(ans[i][j]);
      }
      writeChar('\n');
    }
    return 0;
  }
  return 0;
}