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.
import java.io.*;
import java.util.*;
import java.math.BigInteger;
import java.util.Map.Entry;

import javax.xml.soap.Node;

import static java.lang.Math.*;

public class B extends PrintWriter {

    void add(int[] fenwickTree, int index, int value) {
        for (; index < fenwickTree.length; index = nextIndexFT(index)) {
            fenwickTree[index] += value;
        }
    }

    int sum(int[] fenwickTree, int index) {
        int sum = 0;
        for (; index >= 0; index = prevIndexFT(index)) {
            sum += fenwickTree[index];
        }
        return sum;
    }

    int sum(int[] fenwickTree, int from, int to) {
        return sum(fenwickTree, to) - sum(fenwickTree, from - 1);
    }

    int nextIndexFT(int index) {
        return index | (index + 1);
    }

    int prevIndexFT(int index) {
        return (index & (index + 1)) - 1;
    }

    void run() {
        int n = nextInt();

        int m = n;
        int l = 0, r = 3 * n + 3;

        int[] ft = new int[r + 1];

        int[] p = nextArray(n);

        long sum = 0;
        for (int i = 0; i < n; i++) {
            int diff = p[i] - i - 1;
            add(ft, diff + m, +1);
            sum += abs(diff);
        }

        long ans = sum;
        int off = 0;

        for (int step = 1; step < n; step++) {
            int i = n - step;

            int oldDiff = p[i] - i - 1;
            add(ft, oldDiff + m, -1);

            sum -= abs(p[i] - n);
            sum += abs(p[i] - 1);

            sum += sum(ft, l, step - 1 + m);
            sum -= sum(ft, step + m, r);

            if (sum < ans) {
                ans = sum;
                off = step;
            }

            int newDiff = p[i] - 1 + step;
            add(ft, newDiff + m, +1);
        }

        print(ans);
        print(' ');
        print(off);

    }

    boolean skip() {
        while (hasNext()) {
            next();
        }
        return true;
    }

    int[][] nextMatrix(int n, int m) {
        int[][] matrix = new int[n][m];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                matrix[i][j] = nextInt();
        return matrix;
    }

    String next() {
        while (!tokenizer.hasMoreTokens())
            tokenizer = new StringTokenizer(nextLine());
        return tokenizer.nextToken();
    }

    boolean hasNext() {
        while (!tokenizer.hasMoreTokens()) {
            String line = nextLine();
            if (line == null) {
                return false;
            }
            tokenizer = new StringTokenizer(line);
        }
        return true;
    }

    int[] nextArray(int n) {
        int[] array = new int[n];
        for (int i = 0; i < n; i++) {
            array[i] = nextInt();
        }
        return array;
    }

    int nextInt() {
        return Integer.parseInt(next());
    }

    long nextLong() {
        return Long.parseLong(next());
    }

    double nextDouble() {
        return Double.parseDouble(next());
    }

    String nextLine() {
        try {
            return reader.readLine();
        } catch (IOException err) {
            return null;
        }
    }

    public B(OutputStream outputStream) {
        super(outputStream);
    }

    static BufferedReader reader;
    static StringTokenizer tokenizer = new StringTokenizer("");
    static Random rnd = new Random();
    static boolean OJ;

    public static void main(String[] args) throws IOException {
        OJ = System.getProperty("ONLINE_JUDGE") != null;
        B solution = new B(System.out);
        if (OJ) {
            reader = new BufferedReader(new InputStreamReader(System.in));
            solution.run();
        } else {
            reader = new BufferedReader(new FileReader(new File(B.class.getName() + ".txt")));
            long timeout = System.currentTimeMillis();
            while (solution.hasNext()) {
                solution.run();
                solution.println();
                solution.println("----------------------------------");
            }
            solution.println("time: " + (System.currentTimeMillis() - timeout));
        }
        solution.close();
        reader.close();
    }
}