One day, Twilight Sparkle is interested in how to sort a sequence of integers a1, a2, ..., an in non-decreasing order. Being a young unicorn, the only operation she can perform is a unit shift. That is, she can move the last element of the sequence to its beginning:

a1, a2, ..., an → an, a1, a2, ..., an - 1.

Help Twilight Sparkle to calculate: what is the minimum number of operations that she needs to sort the sequence?

Input

The first line contains an integer n (2 ≤ n ≤ 105). The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 105).

Output

If it's impossible to sort the sequence output -1. Otherwise output the minimum number of operations Twilight Sparkle needs to sort it.

Examples

Input

2
2 1


Output

1


Input

3
1 3 2


Output

-1


Input

2
1 2


Output

0
#include <bits/stdc++.h>
long long exp(long long x, long long y, long long p) {
  long long res = 1;
  while (y) {
    if (y % 2) res = (res * x % p) % p;
    x = (x * x) % p;
    y /= 2;
  }
  return res;
}
long long expo(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y % 2) res = (res * x % ((long long)1e9 + 7)) % ((long long)1e9 + 7);
    x = (x * x) % ((long long)1e9 + 7);
    y /= 2;
  }
  return res;
}
long long sub(long long a, long long b) {
  return (a % ((long long)1e9 + 7) - b % ((long long)1e9 + 7) +
          ((long long)1e9 + 7)) %
         ((long long)1e9 + 7);
}
long long mul(long long a, long long b) {
  return ((a % ((long long)1e9 + 7)) * (b % ((long long)1e9 + 7)) +
          ((long long)1e9 + 7)) %
         ((long long)1e9 + 7);
}
long long inv(long long x) { return expo(x, ((long long)1e9 + 7) - 2); }
using namespace std;
void solve() {
  long long n;
  cin >> n;
  long long f = 0;
  long long b[n];
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  if (is_sorted(a, a + n)) {
    cout << "0" << endl;
    return;
  }
  long long cnt = 0;
  for (long long i = 1; i < n; i++) {
    if (a[i] < a[i - 1]) {
      cnt++;
      f = i;
    }
  }
  if (a[0] < a[n - 1]) cnt++;
  if (cnt < 2) {
    cout << n - f << endl;
    return;
  }
  cout << -1 << endl;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  return 0;
}
