Mister B and PR Shifts——思维题

题目

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 (sum_{i=1}^{i=n}left | p[i]-i ight |).

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 (p_1, p_2, ... p_n),
(k = 1): shift (p_n, p_1, ... p_{n - 1}),
...,
(k = n - 1): shift (p_2, p_3, ... p_n, p_1).

Input

First line contains single integer (n (2 ≤ n ≤ 10^6)) — the length of the permutation.

The second line contains (n) space-separated integers (p_1, p_2, ..., p_n (1 ≤ p_i ≤ 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 1

3
1 2 3

Output 1

0 0

Input 2

3
2 3 1

Output 2

0 1

Input 3

3
3 2 1

Output 3

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.

题解

解题思路

暴力枚举的话时间复杂度是(n^2)的，对于1e6的数据一定是超时的

(l)统计的是(a_i >= i)的数量
(r)统计的是(a_i < i)的数量
(s_i)统计(a_x - x == i)的数量
每次循环都把最后一位拿到第一位，其他的向后推一位
这样(l)就减去(s_i),(r)加上(s_i)

代码

#include <cstdio>
#include <cmath>
using namespace std;
const int N = 2e6+5;
int n, a[N], s[N], l, r, h;
long long sum, ans;
int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        sum += abs(a[i] - i);
        if (a[i] >= i) l++, s[a[i]-i]++;
        else r++;
    }
    ans = sum; h = 0;
    for(int i = 0; i < n; i++) {
        l -= s[i]; r += s[i];
        sum += r - l - 1;
        sum += 2 * a[n-i] - n - 1;
        l++, r--;
        s[i+a[n-i]]++;
        if (ans > sum) ans = sum, h = i + 1;
    }
    printf("%lld %d", ans, h);
    return 0;
}