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 int MAXN = 1e5 + 5;
const int INF = 2e9;
const int MOD = 1e9 + 7;
long long arr[MAXN], T[MAXN * 4];
long long gcd(long long A, long long B) {
  while (B != 0) {
    A %= B;
    swap(A, B);
  }
  return A;
}
void build(int v, int tl, int tr) {
  if (tl == tr) {
    T[v] = arr[tl];
    return;
  }
  int mid = (tl + tr) >> 1;
  build(v * 2, tl, mid);
  build(v * 2 + 1, mid + 1, tr);
  T[v] = gcd(T[v * 2], T[v * 2 + 1]);
}
void upd(int v, int tl, int tr, int pos, long long new_val) {
  if (tl == tr) {
    T[v] = new_val;
    return;
  }
  int mid = (tl + tr) >> 1;
  if (pos <= mid)
    upd(v * 2, tl, mid, pos, new_val);
  else
    upd(v * 2 + 1, mid + 1, tr, pos, new_val);
  T[v] = gcd(T[v * 2], T[v * 2 + 1]);
}
int main() {
  int N = 0;
  scanf("%d", &N);
  long long cur_gcd = 0, RES = 0;
  for (int i = 1; i <= N; i++) scanf("%lld", &arr[i]);
  build(1, 1, N);
  if (T[1] > 1) return printf("YES\n0"), 0;
  for (int i = 1; i + 1 <= N; i++) {
    if (arr[i] % 2 == 1 && arr[i + 1] % 2 == 1) {
      long long l = arr[i] - arr[i + 1], r = arr[i] + arr[i + 1];
      arr[i] = l, arr[i + 1] = r, RES++;
      upd(1, 1, N, i, l);
      upd(1, 1, N, i + 1, r);
    }
    if (T[1] > 1) return printf("YES\n%lld", RES), 0;
  }
  for (int i = 1; i + 1 <= N; i++) {
    for (int j = 1; j <= 2; j++, RES++) {
      if (T[1] > 1 || gcd(arr[i], arr[i + 1]) % 2 == 0) break;
      long long l = arr[i] - arr[i + 1], r = arr[i] + arr[i + 1];
      arr[i] = l, arr[i + 1] = r;
      upd(1, 1, N, i, l);
      upd(1, 1, N, i + 1, r);
    }
  }
  printf("YES\n");
  printf("%lld", RES);
  return 0;
}
