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;
const int maxn = 1000500;
inline int g() {
  int x;
  cin >> x;
  return x;
}
inline long long gl() {
  long long x;
  cin >> x;
  return x;
}
inline char gc() {
  char x;
  cin >> x;
  return x;
}
inline string gs() {
  string s;
  cin >> s;
  return s;
}
template <typename T>
inline vector<T> gv(int len) {
  vector<T> a(len);
  for (auto& it : a) cin >> it;
  return a;
}
inline vector<int> gv(int len) {
  vector<int> a(len);
  for (auto& it : a) cin >> it;
  return a;
}
inline vector<pair<int, int>> gv_andPairCnt(int len) {
  vector<pair<int, int>> a(len);
  int i = 1;
  for (auto& it : a) {
    it.second = i;
    i++;
    cin >> it.first;
  }
  return a;
}
double power_log(double power, double val) { return log(val) / log(power); }
template <typename T>
T sqr(T x) {
  return x * x;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  for (auto it : v) os << it << ", ";
  os << "\n";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const deque<T>& v) {
  os << "deque:{";
  for (auto it : v) os << it << " ";
  os << "}\n";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  os << "set:{";
  for (auto it : v) os << it << " ";
  os << "}\n";
  return os;
}
template <typename T, typename H>
inline vector<pair<T, H>> gv_andBiPair(int len) {
  vector<pair<T, H>> a(len);
  for (auto& it : a) {
    cin >> it.first >> it.second;
  }
  return a;
}
template <typename T>
inline set<T> vector_to_set(const vector<T>& vec) {
  set<T> s;
  for (auto& it : vec) s.insert(it);
  return s;
}
const long long md = 1e9 + 7;
vector<int> arr, a;
int dp[101][131074], back[101][131074], fact[101];
auto prime = [](int x) {
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      return false;
    }
  }
  return true;
};
int rec(int v, int mask) {
  if (v == a.size()) {
    return 0;
  }
  if (dp[v][mask] != -1) {
    return dp[v][mask];
  }
  int& x = dp[v][mask];
  x = abs(1 - a[v]) + rec(v + 1, mask);
  back[v][mask] = 1;
  for (int i = 2; i <= 60; i++) {
    if ((fact[i] & mask) > 0) {
      continue;
    }
    int X = abs(i - a[v]) + rec(v + 1, mask | fact[i]);
    if (x > X) {
      x = X;
      back[v][mask] = i;
    }
  }
  return x;
}
void ans(int v, int mask) {
  if (v == a.size()) {
    return;
  }
  cout << back[v][mask] << " ";
  ans(v + 1, mask | fact[back[v][mask]]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  memset(dp, -1, sizeof(dp));
  int n = g();
  a = gv(n);
  for (int i = 2; i <= 60; i++) {
    if (prime(i)) {
      arr.push_back(i);
    }
  }
  for (int i = 1; i <= 60; i++) {
    for (int j = 0; j < arr.size(); j++) {
      if (i % arr[j] == 0) {
        fact[i] |= (1 << j);
      }
    }
  }
  rec(0, 0);
  ans(0, 0);
}