819_B. Mister B and PR Shifts
Some time ago Mister B detected a strange signal from the space, which he started to study.

After some transformation the signal turned out to be a permutation p of length n or its cyclic shift. For the further investigation Mister B need some basis, that's why he decided to choose cyclic shift of this permutation which has the minimum possible deviation.

Let's define the deviation of a permutation p as <image>.

Find a cyclic shift of permutation p with minimum possible deviation. If there are multiple solutions, print any of them.

Let's denote id k (0 ≤ k < n) of a cyclic shift of permutation p as the number of right shifts needed to reach this shift, for example:

  * k = 0: shift p1, p2, ... pn, 
  * k = 1: shift pn, p1, ... pn - 1, 
  * ..., 
  * k = n - 1: shift p2, p3, ... pn, p1. 

Input

First line contains single integer n (2 ≤ n ≤ 106) — the length of the permutation.

The second line contains n space-separated integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the elements of the permutation. It is guaranteed that all elements are distinct.

Output

Print two integers: the minimum deviation of cyclic shifts of permutation p and the id of such shift. If there are multiple solutions, print any of them.

Examples

Input

3
1 2 3


Output

0 0


Input

3
2 3 1


Output

0 1


Input

3
3 2 1


Output

2 1

Note

In the first sample test the given permutation p is the identity permutation, that's why its deviation equals to 0, the shift id equals to 0 as well.

In the second sample test the deviation of p equals to 4, the deviation of the 1-st cyclic shift (1, 2, 3) equals to 0, the deviation of the 2-nd cyclic shift (3, 1, 2) equals to 4, the optimal is the 1-st cyclic shift.

In the third sample test the deviation of p equals to 4, the deviation of the 1-st cyclic shift (1, 3, 2) equals to 2, the deviation of the 2-nd cyclic shift (2, 1, 3) also equals to 2, so the optimal are both 1-st and 2-nd cyclic shifts.
```python
def main():
	n = int(input())
	data = input().split()

	#print(str(n) + " " + str(data))

	data = list(map(lambda x: int(x), data))
	res = 0
	ires = 0
	neg = 0
	when = [0] * n

	for i in range(n):
		data[i] = i + 1 - data[i]
		res += abs(data[i])
		if data[i] <= 0:
			neg += 1
		a = -data[i]
		if a < 0:
			a = a + n
		when[a] += 1
	#print(when)
	
	ares = res	
	#print(str(res) + " " + str(ires) + " " + str(neg))
	for i in range(n):
		neg -= when[i]
		ares -= neg
		ares += (n - neg)
		x = data[n - i - 1] + i + 1
		ares -= x
		ares += n - x
		#print(str(res) + " " + str(ires) + " " + str(ares) + " " + str(i) + " " + str(neg))
		neg += 1
		if ares < res:
			res = ares
			ires = i + 1

	print(str(res) + " " + str(ires))
	
main()

```