Problem: Princess Twilight went to Celestia and Luna's old castle to research the chest from the Elements of Harmony.

<image>

A sequence of positive integers bi is harmony if and only if for every two elements of the sequence their greatest common divisor equals 1. According to an ancient book, the key of the chest is a harmony sequence bi which minimizes the following expression:

<image>

You are given sequence ai, help Princess Twilight to find the key.

Input

The first line contains an integer n (1 ≤ n ≤ 100) — the number of elements of the sequences a and b. The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 30).

Output

Output the key — sequence bi that minimizes the sum described above. If there are multiple optimal sequences, you can output any of them.

Examples

Input

5
1 1 1 1 1


Output

1 1 1 1 1 

Input

5
1 6 4 2 8


Output

1 5 3 1 8 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<vector<long long> > matmul(vector<vector<long long> > &a,
                                  vector<vector<long long> > &b) {
  int i, j, k, n = a.size();
  vector<vector<long long> > ans(n, vector<long long>(n));
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      long long temp = 0;
      for (k = 0; k < n; k++)
        temp = (temp + a[i][k] * b[k][j]) % (1000000007LL);
      ans[i][j] = temp;
    }
  return ans;
}
vector<vector<long long> > matpwr(vector<vector<long long> > &a, long long p) {
  int n = a.size();
  vector<vector<long long> > ans(n, vector<long long>(n));
  for (int i = 0; i < n; i++) ans[i][i] = 1;
  while (p) {
    if (p & 1) ans = matmul(ans, a);
    a = matmul(a, a);
    p >>= 1;
  }
  return ans;
}
long long pwr(long long base, long long p, long long mod = (1000000007LL)) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    p /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long invMod(long long n, long long mod = (1000000007LL)) {
  return pwr(n, mod - 2);
}
int primes[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53};
int n, arr[105], maskVal[60], DP[105][1 << 17], ele[105][1 << 17],
    go[105][1 << 17], p = 16;
vector<int> valid[(1 << 17) + 2];
int dp(int pos, int mask) {
  if (pos == n) return 0;
  if (DP[pos][mask] != -1) return DP[pos][mask];
  int ans = (100000005);
  for (int i = 0; i < (int)valid[mask].size(); i++) {
    int val = valid[mask][i];
    if (abs(val - arr[pos]) + dp(pos + 1, maskVal[val] | mask) < ans) {
      ans = abs(val - arr[pos]) + dp(pos + 1, maskVal[val] | mask);
      ele[pos][mask] = val;
      go[pos][mask] = maskVal[val] | mask;
    }
  }
  return DP[pos][mask] = ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> arr[i];
  for (int mask = 0; mask < (1 << p); mask++)
    for (int i = 1; i <= 58; i++) {
      bool invalid = false;
      for (int j = 0; j < p; j++)
        if (i % primes[j] == 0 && ((mask >> j) & 1)) invalid = true;
      if (!invalid) valid[mask].push_back(i);
    }
  for (int i = 1; i <= 58; i++) {
    int mask = 0;
    for (int j = 0; j < p; j++)
      if (i % primes[j] == 0) mask |= (1 << j);
    maskVal[i] = mask;
  }
  memset(DP, -1, sizeof(DP));
  memset(ele, -1, sizeof(ele));
  memset(go, -1, sizeof(go));
  dp(0, 0);
  int i = 0, mask = 0;
  while (i < n) {
    assert(ele[i][mask] != -1);
    assert(go[i][mask] != -1);
    cout << ele[i][mask] << " ";
    mask = go[i][mask];
    i++;
  }
  return 0;
}