B. Little Pony and Sort by Shift time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

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, ..., a

n - 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.

Sample test(s) input

2
2 1

output

1

input

3
1 3 2

output

-1

input

2
1 2

output

0

題意：一串數字序列，可以將最後的移到最前面去。問一個打亂的序列需要移動幾次才能變為一個由小到大的序列，如果不能，則輸出-1. 其實本題只需要看是否為1個或2個遞增的序列，並且當有2個遞增的序列時需要滿足遞增序列的末點要比起點小或相等。

#include<stdio.h>
int main()
{
    int n, i, j, a[100010];

    while(~scanf("%d", &n))
    {
        for(i=0; i<n; i++)
            scanf("%d", a+i);
        int flag = 0, index = 0;
        for(i=1; i<n; i++)
            if(a[i] < a[i-1]) flag++, index = i;
        if(flag > 1 || (flag > 0 && a[n-1] > a[0]))
        {
            printf("-1\n");
            continue;
        }
        for(i=1; i<n; i++)
            if(a[i] < a[i-1]) break;
        if(i == n) index = n;
        printf("%d\n", n-index);
    }
    return 0;
}