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

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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:

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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;
int SET(int n, int pos) { return n = n | (1 << pos); }
int RESET(int n, int pos) { return n = n & ~(1 << pos); }
int CHECK(int n, int pos) { return (bool)(n & (1 << pos)); }
int str2int(string s) {
  stringstream ss(s);
  int x;
  ss >> x;
  return x;
}
string int2str(int a) {
  stringstream ss;
  ss << a;
  string str = ss.str();
  return str;
}
string char2str(char a) {
  stringstream ss;
  ss << a;
  string str = ss.str();
  return str;
}
long long int bigMod(long long int n, long long int power, long long int MOD) {
  if (power == 0) return 1;
  if (power % 2 == 0) {
    long long int ret = bigMod(n, power / 2, MOD);
    return ((ret % MOD) * (ret % MOD)) % MOD;
  } else
    return ((n % MOD) * (bigMod(n, power - 1, MOD) % MOD)) % MOD;
}
long long int modInverse(long long int n, long long int MOD) {
  return bigMod(n, MOD - 2, MOD);
}
int POW(int x, int y) {
  int res = 1;
  for (; y;) {
    if ((y & 1)) {
      res *= x;
    }
    x *= x;
    y >>= 1;
  }
  return res;
}
int inverse(int x) {
  double p = ((double)1.0) / x;
  return (p) + 1e-9;
}
int gcd(int a, int b) {
  while (b) b ^= a ^= b ^= a %= b;
  return a;
}
int nC2(int n) { return n * (n - 1) / 2; }
long long int MOD(long long int n, long long int mod) {
  if (n >= 0)
    return n % mod;
  else if (-n == mod)
    return 0;
  else
    return mod + (n % mod);
}
int n, data[105], Mask[65], dp[105][(1 << 18) + 1], path[105][(1 << 18) + 1];
bool isPrime[65];
vector<int> prime;
void seive_N_logN(int N) {
  memset(isPrime, true, sizeof isPrime);
  prime.clear();
  isPrime[1] = false;
  for (int i = 4; i <= N; i = i + 2) isPrime[i] = false;
  for (int i = 3; i * i <= N; i = i + 2) {
    if (isPrime[i]) {
      for (int j = i * i; j <= N; j += i) isPrime[j] = false;
    }
  }
  for (int i = 1; i < N; i++)
    if (isPrime[i]) prime.push_back(i);
  for (int i = 1; i <= 60; i++) {
    for (int j = 0; j < prime.size(); j++) {
      if (i % prime[j] == 0) Mask[i] = SET(Mask[i], j);
    }
  }
}
int call(int i, int mask) {
  if (i >= n) {
    return 0;
  }
  int &ret = dp[i][mask];
  if (ret != -1) return ret;
  ret = 1000000000;
  for (int j = 1; j <= 60; j++) {
    if (mask & Mask[j]) continue;
    int xx = call(i + 1, mask | Mask[j]) + abs(data[i] - j);
    if (xx < ret) {
      ret = xx;
      path[i][mask] = j;
    }
  }
  return ret;
}
int main() {
  seive_N_logN(60);
  int t, cas = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &data[i]);
  }
  memset(dp, -1, sizeof(dp));
  call(0, 0);
  int mask = 0;
  for (int i = 0; i < n; i++) {
    int ans = path[i][mask];
    printf("%d ", ans);
    mask |= Mask[ans];
  }
  printf("\n");
  return 0;
}