Problem: You are given an array of n integers. You need to split all integers into two groups so that the GCD of all integers in the first group is equal to one and the GCD of all integers in the second group is equal to one.

The GCD of a group of integers is the largest non-negative integer that divides all the integers in the group.

Both groups have to be non-empty.

Input

The first line contains a single integer n (2 ≤ n ≤ 10^5).

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the array.

Output

In the first line print "YES" (without quotes), if it is possible to split the integers into two groups as required, and "NO" (without quotes) otherwise.

If it is possible to split the integers, in the second line print n integers, where the i-th integer is equal to 1 if the integer a_i should be in the first group, and 2 otherwise.

If there are multiple solutions, print any.

Examples

Input


4
2 3 6 7


Output


YES
2 2 1 1 


Input


5
6 15 35 77 22


Output


YES
2 1 2 1 1 


Input


5
6 10 15 1000 75


Output


NO

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long MODpow(long long input, long long n, long long difference) {
  long long res = 1, y = input;
  while (n > 0) {
    if (n & 1) res = (res * y) % difference;
    y = (y * y) % difference;
    n /= 2;
  }
  return res % difference;
}
template <class T>
T _abs(T n) {
  return (n < 0 ? -n : n);
}
template <class T>
T _max(T input, T b) {
  return (!(input < b) ? input : b);
}
template <class T>
T _min(T input, T b) {
  return (input < b ? input : b);
}
template <class T>
T gcd(T input, T b) {
  return (b != 0 ? gcd<T>(b, input % b) : input);
}
template <class T>
T lcm(T input, T b) {
  return (input / gcd<T>(input, b) * b);
}
inline int fi() {
  int n = 0;
  int ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9')
    n = (n << 3) + (n << 1) + ch - '0', ch = getchar();
  return n;
}
inline void fo(int input) {
  char snum[20];
  int iterator = 0;
  do {
    snum[iterator++] = input % 10 + 48;
    input = input / 10;
  } while (input != 0);
  iterator = iterator - 1;
  while (iterator >= 0) putchar(snum[iterator--]);
  putchar('\n');
}
int MOD = 1e9 + 7, INF = 1e9;
long long INFLL = 1e18;
long long unsigned INFLLU = 18446744073709551615ULL;
long double EPS = 1e-12;
inline int inv(int input, int p = MOD) { return MODpow(input, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline void addMOD(int& input, int val, int p = MOD) {
  if ((input = (input + val)) >= p) input -= p;
}
inline void subMOD(int& input, int val, int p = MOD) {
  if ((input = (input - val)) < 0) input += p;
}
const int N = 1e5 + 5;
int a[N], p[N], ans[N];
int main() {
  srand(19260817);
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) p[i] = i;
  int T = 100;
  while (T--) {
    random_shuffle(p + 1, p + n + 1);
    int x = 0, y = 0, tmp;
    for (int i = 1; i <= n; i++) {
      if ((x != 1) && ((tmp = gcd(x, a[p[i]])) != x))
        x = tmp, ans[p[i]] = 1;
      else
        y = gcd(y, a[p[i]]), ans[p[i]] = 2;
    }
    if (x == 1 && y == 1) {
      puts("YES");
      for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
      return 0;
    }
  }
  puts("NO");
  return 0;
}