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;
bool isprime[1000001];
void sieve() {
  for (long long i = 0; i <= 1000001; i++) {
    isprime[i] = true;
  }
  isprime[0] = false;
  isprime[1] = false;
  for (long long i = 2; i * i <= 1000001; i++) {
    if (isprime[i] == true) {
      for (long long j = i * i; j <= 1000001; j = j + i) {
        isprime[j] = false;
      }
    }
  }
}
long long gcd(long long a, long long b) {
  if (a < b) {
    swap(a, b);
  }
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
bool comparen(pair<long long, long long> a, pair<long long, long long> b) {
  return a.second < b.second;
}
bool comparem(pair<long long, long long> a, pair<long long, long long> b) {
  return a.first > b.first;
}
void remove(long long n, char x, string s) { ; }
void solve() {
  long long t = 1;
  while (t--) {
    long long n;
    cin >> n;
    long long a[n];
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long i = 0;
    while (i < n - 1 && a[i] <= a[i + 1]) {
      i++;
    }
    long long k = i + 1;
    i++;
    while (i < n - 1 && a[i] <= a[i + 1]) {
      i++;
    }
    if (i == n) {
      cout << 0 << endl;
    } else if (i != n - 1 || a[0] < a[n - 1]) {
      cout << -1 << endl;
    } else {
      cout << n - k << endl;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  return 0;
}
