Mike has a sequence A = [a1, a2, ..., an] of length n. He considers the sequence B = [b1, b2, ..., bn] beautiful if the gcd of all its elements is bigger than 1, i.e. <image>. 

Mike wants to change his sequence in order to make it beautiful. In one move he can choose an index i (1 ≤ i < n), delete numbers ai, ai + 1 and put numbers ai - ai + 1, ai + ai + 1 in their place instead, in this order. He wants perform as few operations as possible. Find the minimal number of operations to make sequence A beautiful if it's possible, or tell him that it is impossible to do so.

<image> is the biggest non-negative number d such that d divides bi for every i (1 ≤ i ≤ n).

Input

The first line contains a single integer n (2 ≤ n ≤ 100 000) — length of sequence A.

The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — elements of sequence A.

Output

Output on the first line "YES" (without quotes) if it is possible to make sequence A beautiful by performing operations described above, and "NO" (without quotes) otherwise.

If the answer was "YES", output the minimal number of moves needed to make sequence A beautiful.

Examples

Input

2
1 1


Output

YES
1


Input

3
6 2 4


Output

YES
0


Input

2
1 3


Output

YES
1

Note

In the first example you can simply make one move to obtain sequence [0, 2] with <image>.

In the second example the gcd of the sequence is already greater than 1. 
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1000000007;
const long long MAXN = 1000001;
bool prime[1000001];
void SieveOfEratosthenes() {
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= 1000000; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= 1000000; i += p) prime[i] = false;
    }
  }
}
long long bs(vector<long long> a, long long x, long long n) {
  int l = 0;
  int h = n - 1;
  int m;
  while (l <= h) {
    m = (l + h) / 2;
    if (a[m] > x)
      h = m - 1;
    else if (a[m] < x)
      l = m + 1;
    else
      return m;
  }
  if (a[m] >= x) m--;
  return m;
}
long long power(long long x, long long y) {
  long long p = INF;
  int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long mex(vector<long long> a, int n) {
  sort(a.begin(), a.end());
  int j = a[0];
  if (j != 0)
    return 0;
  else {
    for (int i = 1; i < n; i++) {
      if (a[i] != a[i - 1]) {
        if (a[i] != a[i - 1] + 1) {
          return j + 1;
        } else
          j++;
      }
    }
    return j + 1;
  }
}
int spf[MAXN];
void sieve() {
  spf[1] = 1;
  for (int i = 2; i < MAXN; i++) spf[i] = i;
  for (int i = 4; i < MAXN; i += 2) spf[i] = 2;
  for (int i = 3; i * i < MAXN; i++) {
    if (spf[i] == i) {
      for (int j = i * i; j < MAXN; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long c(long long m, long long n) {
  long long ans = 1;
  for (int i = 1; i <= n; i++) {
    ans = (ans * (m - i + 1)) / i;
  }
  return ans;
}
long long modInverse(long long a, long long m) {
  long long m0 = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
long long pc(long long n) {
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
long long lexo(string a, string b) {
  long long n = a.length();
  for (int i = 0; i < n; i++) {
    if (a[i] < b[i])
      return 1;
    else if (b[i] < a[i])
      return 0;
  }
  return 0;
}
long long mod(long long x) { return (x % INF + INF) % INF; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long test;
  test = 1;
  while (test--) {
    long long n;
    cin >> n;
    long long a[n];
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long ans = 0;
    long long g = 0;
    for (int i = 0; i < n; i++) {
      g = gcd(g, a[i]);
    }
    if (g == 1) {
      for (int i = 0; i < n - 1; i++) {
        if (a[i] % 2) {
          if (a[i + 1] % 2 == 0) {
            ans += 2;
            a[i]++;
          } else {
            a[i]++;
            a[i + 1]++;
            ans++;
            i++;
          }
        }
      }
      if (a[n - 1] % 2) {
        if (a[n - 2] % 2)
          ans++;
        else
          ans += 2;
      }
    }
    cout << "YES"
         << "\n";
    cout << ans << "\n";
  }
}
