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>
using namespace std;
long mod = pow(10, 9) + 7;
int countdivisors(int n) {
  long long count = 0;
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i) {
        count++;
      } else {
        count = count + 2;
      }
    }
  }
  return count;
}
void remove(std::vector<int> &v) {
  auto end = v.end();
  for (auto it = v.begin(); it != end; ++it) {
    end = std::remove(it + 1, end, *it);
  }
  v.erase(end, v.end());
}
bool isPerfectSquare(int x) {
  int s = sqrt(x);
  return (s * s == x);
}
bool isFibonacci(int n) {
  return isPerfectSquare(5 * n * n + 4) || isPerfectSquare(5 * n * n - 4);
}
int digSum(int n) {
  int sum = 0;
  while (n > 0) {
    sum += n % 10;
    n = n / 10;
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  if (is_sorted(v.begin(), v.end())) {
    cout << "0";
  } else {
    int x;
    for (int i = 0; i < n - 1; i++) {
      if (v[i] > v[i + 1]) {
        x = i + 1;
        break;
      }
    }
    vector<int> s;
    for (int i = x; i < n; i++) {
      s.push_back(v[i]);
    }
    for (int i = 0; i < x; i++) {
      s.push_back(v[i]);
    }
    if (is_sorted(s.begin(), s.end())) {
      cout << (n - x);
    } else {
      cout << "-1";
    }
  }
  return 0;
}
