Codeforces Round #259 (Div. 2) b. Little Pony and Sort by Shift

One day, Twilight Sparkle are interested in what to sort a sequence of integers a_{1,? A2,?...,? An} in non-decreasing order. Being A young unicorn, the only operation she can perform is a unit shift. That's, she can move the last element of the sequence to its beginning:

a _{1,? a 2,?...,? a n? →? a n,? a 1,? a 2,?...,? a n?-? 1.}

Help Twilight Sparkle to Calculate:what are 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 containsn integer numbers a_{1,? A2,?...,? An}(1?≤? ) A_{i? ≤?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

Idea: Just start thinking is wrong, only consider the adjacent three digits-! , you can try it out wrong when you try the data.

See how many a[i-1]>a[i], use S to write down, and then use now to write down the current number of subscript. To move (shift) The number is n-now-1, and then look at the value of S, if s==1, indicates that there may be movement to become

# include < iostream >
# include < cstrings >
# include < cstdio >
# include < string >
# include < cmath >
# include < algorithm >
# define LL int
# define 0 x3f3f3f3f inf
Using namespace STD.
Int a, [1000001].
Int main ()
{
Int n, m, I, j; Int k, x, y;
While (~ the scanf (" % d ", & n))
{
For (I = 0; I < n. I++)
{
The scanf (" % d ", & a [I]);
}
Int s = 0, now = 0; // form an initialization habit
For (I = 0; I < n - 1; I++)
{
If (a > [I] a [I + 1))
{
Now = I; < span id = "transmark" > < / span >
S++;
}
}
If (s = = 0)
{
Printf (" 0 \ n ");
The continue;
}
Else if (s = = 1 [0] && a > = a/n - 1)
{
Printf (" % d \ n, n - now - 1);
}
The else
{
Printf (" \ n - 1 ");
}
}
Return 0;
} 

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Codeforces Round #259 (Div. 2) b. Little Pony and Sort by Shift