time limit per test 2 seconds

memory limit per test 256 megabytes

input standard input

output standard output

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 .

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

題意：

有n個數，從1到n亂序排列，定義這n個數的秩序值為∑(a[i]-i) (1<=i<=n)， 你每次將這個陣列向右迴圈移位p次，問p等於多少時，這n個數的秩序值最小？

記錄下每個數是在目標位置的左邊還是右邊，存下所有在目標左邊的數，

cur[i]表示初始陣列中有多少個數在它目標左邊第i個位置上

->注意要存下一開始在目標左邊（包括在目標上）的個數L，以及目標右邊的數的個數r

之後直接模擬右移，對於第i次右移，L = L-cur[i-1],   r = r+cur[i-1]，即當次位移剛好有cur[i-1]個數原本在目標位置左邊及目標上跑到了目標位置的右邊，

每次的秩序值便是初始秩序值-L+r+特判最後一個數移到第一個數所產生的影響

但注意計算之後L還要+1，r還要-1因為最後一個數跑到了第一個

複雜度O(n)

#include<stdio.h>
#include<stdlib.h>
#define LL long long
int p[1000005], cur[2000005];
int main(void)
{
	LL ans, sum;
	int n, L, r, i, temp;
	scanf("%d", &n);
	sum = L = r = temp = 0;
	for(i=1;i<=n;i++)
		scanf("%d", &p[i]);
	for(i=1;i<=n;i++)
	{
		sum += abs(p[i]-i);
		if(p[i]>=i)  L++, cur[p[i]-i]++;
		else  r++;
	}
	ans = sum;
	for(i=0;i<n-1;i++)
	{
		L -= cur[i]; r += cur[i];
		sum = sum-L+r-abs(p[n-i]-n-1)+p[n-i]-1;
		cur[p[n-i]+i]++;
		L++, r--;
		if(sum<ans)
			ans = sum, temp = i+1;
	}
	printf("%lld %d\n", ans, temp);
	return 0;
}