Problem: In this task you have to write a program dealing with nonograms on fields no larger than 5 × 20.

Simplified nonogram is a task where you have to build such field (each cell is either white or black) that satisfies the given information about rows and columns. For each row and each column the number of contiguous black segments is specified. 

For example if size of the field is n = 3, m = 5, аnd numbers of contiguous black segments in rows are: [2, 3, 2] and in columns are: [1, 0, 1, 2, 1] then the solution may look like:

<image>

It is guaranteed that on each test in the testset there exists at least one solution.

Input

In the first line there follow two integers n, m (1 ≤ n ≤ 5, 1 ≤ m ≤ 20) — number of rows and number of columns respectively.

Second line contains n integers a1, a2, ..., an where ai is the number of contiguous black segments in i-th row of the field. 

Similarly, third line contains m integers b1, b2, ..., bm where bi is the number of contiguous black segments in the i-th column of the field.

It is guaranteed that there exists at least one solution.

Output

Output any possible solution. Output should consist of n lines each containing m characters. Denote white cell as "." and black cell as "*".

Examples

Input

3 5
2 3 2
1 0 1 2 1


Output

*.**.
*.*.*
*..**

Input

3 3
2 1 2
2 1 2


Output

*.*
.*.
*.*


Input

3 3
1 0 1
2 2 2


Output

***
...
***

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void read(int &x) {
  int v = 0, f = 1;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-')
    f = -1;
  else
    v = v * 10 + c - '0';
  while (isdigit(c = getchar())) v = v * 10 + c - '0';
  x = v * f;
}
void read(long long &x) {
  long long v = 0ll, f = 1ll;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-')
    f = -1;
  else
    v = v * 10 + c - '0';
  while (isdigit(c = getchar())) v = v * 10 + c - '0';
  x = v * f;
}
void readc(char &x) {
  char c;
  while ((c = getchar()) == ' ')
    ;
  x = c;
}
void writes(string s) { puts(s.c_str()); }
void writeln() { writes(""); }
void writei(int x) {
  if (x < 0) {
    putchar('-');
    x = abs(x);
  }
  if (!x) putchar('0');
  char a[25];
  int top = 0;
  while (x) {
    a[++top] = (x % 10) + '0';
    x /= 10;
  }
  while (top) {
    putchar(a[top]);
    top--;
  }
}
void writell(long long x) {
  if (x < 0) {
    putchar('-');
    x = abs(x);
  }
  if (!x) putchar('0');
  char a[25];
  int top = 0;
  while (x) {
    a[++top] = (x % 10) + '0';
    x /= 10;
  }
  while (top) {
    putchar(a[top]);
    top--;
  }
}
inline long long inc(int &x) { return ++x; }
inline long long inc(long long &x) { return ++x; }
inline long long inc(int &x, long long y) { return x += y; }
inline long long inc(long long &x, long long y) { return x += y; }
inline double inc(double &x, double y) { return x += y; }
inline long long dec(int &x) { return --x; }
inline long long dec(long long &x) { return --x; }
inline long long dec(int &x, long long y) { return x -= y; }
inline long long dec(long long &x, long long y) { return x -= y; }
inline double dec(double &x, double y) { return x -= y; }
inline long long mul(int &x) { return x = ((long long)x) * x; }
inline long long mul(long long &x) { return x = x * x; }
inline long long mul(int &x, long long y) { return x *= y; }
inline long long mul(long long &x, long long y) { return x *= y; }
inline double mul(double &x, double y) { return x *= y; }
inline long long divi(const int &x) {
  long long ans, l, r, mid;
  ans = 0;
  l = 0;
  r = 0x3fffffff;
  while (l < r) {
    mid = (l + r) / 2;
    if (mid * mid <= x) {
      ans = mid;
      l = mid + 1;
    } else
      r = mid;
  }
  return ans;
}
inline long long divi(const long long &x) {
  long long ans, l, r, mid;
  ans = 0;
  l = 0;
  r = 0x3fffffff;
  while (l < r) {
    mid = (l + r) / 2;
    if (mid * mid <= x) {
      ans = mid;
      l = mid + 1;
    } else
      r = mid;
  }
  return ans;
}
inline long long divi(int &x, long long y) { return x /= y; }
inline long long divi(long long &x, long long y) { return x /= y; }
inline double divi(double &x, double y) { return x /= y; }
inline long long mod(int &x, long long y) { return x %= y; }
inline long long mod(long long &x, long long y) { return x %= y; }
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fdelete-null-pointer-checks")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC target("avx")
int n, m, i, j, k, l, r[25], c[25], rr;
short dp[21][161051][32];
char ans[25][25];
int gt(int x) {
  int i, s = 0;
  for (i = 0; i < 5; i++) {
    s *= 10;
    if ((x >> i) & 1) s++;
  }
  return s;
}
void print(int x, int y, int z) {
  if (x == 0) return;
  int i;
  for (i = 0; i < 5; i++) {
    if ((1 << i) & z)
      ans[i][x] = '*';
    else
      ans[i][x] = '.';
  }
  print(x - 1, y - gt(z & (31 ^ dp[x][y][z])), dp[x][y][z]);
}
int kk(int x) {
  int s = 0, i, l = 0;
  for (i = 0; i < 5; i++) {
    if ((x >> i) & 1) {
      if (l == 1) continue;
      s++;
      l = 1;
    } else
      l = 0;
  }
  return s;
}
int main() {
  cin >> n >> m;
  if ((1) <= ((5)))
    for (((i)) = (1); ((i)) <= ((5)); ((i))++) {
      if (i <= n) cin >> r[i];
      rr = rr * 10 + r[i];
    }
  if ((1) <= ((m)))
    for (((i)) = (1); ((i)) <= ((m)); ((i))++) cin >> c[i];
  memset(dp, -1, sizeof(dp));
  dp[0][0][0] = 0;
  if ((0) <= (m - 1))
    for ((i) = (0); (i) <= (m - 1); (i)++) {
      if ((0) <= (161050))
        for ((j) = (0); (j) <= (161050); (j)++) {
          if ((0) <= (31))
            for ((k) = (0); (k) <= (31); (k)++) {
              if (dp[i][j][k] == -1) continue;
              if ((0) <= (31))
                for ((l) = (0); (l) <= (31); (l)++)
                  if (kk(l) == c[i + 1]) {
                    dp[i + 1][j + gt(l & (31 ^ k))][l] = k;
                  }
            }
        }
    }
  if ((0) <= (31))
    for ((k) = (0); (k) <= (31); (k)++) {
      if (dp[m][rr][k] != -1) {
        print(m, rr, k);
        if ((0) <= (n - 1))
          for ((i) = (0); (i) <= (n - 1); (i)++) {
            if ((1) <= ((m)))
              for (((j)) = (1); ((j)) <= ((m)); ((j))++) cout << ans[i][j];
            cout << endl;
          }
        return 0;
      }
    }
  return 0;
}