#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using PII = pair<int, int>;
using VI = vector<int>;
class SegmentTree {
  int sz;
  vector<ll> add, dif;

 public:
  SegmentTree(int n) {
    for (sz = 1; sz < n; sz <<= 1)
      ;
    add.resize(2 * sz);
    dif.resize(2 * sz);
  }
  void update(int l, int r, ll a, ll d, int node = 1, int x = 0, int y = -1) {
    if (y == -1) y = sz;
    if (l >= y || r <= x) return;
    if (l <= x && y <= r) {
      add[node] += a + (x - l) * d;
      dif[node] += d;
      return;
    }
    update(l, r, a, d, 2 * node, x, (x + y) / 2);
    update(l, r, a, d, 2 * node + 1, (x + y) / 2, y);
  }
  ll query(int i, int node = 1, int x = 0, int y = -1) {
    if (y == -1) y = sz;
    if (i >= y || i < x) return 0;
    if (y - x == 1) return add[node];
    ll res = add[node] + (i - x) * dif[node];
    int z = (x + y) / 2;
    if (i < z)
      res += query(i, 2 * node, x, z);
    else
      res += query(i, 2 * node + 1, z, y);
    return res;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  SegmentTree tree(n + 1);
  for (int i = 1; i <= n; i++) {
    int pi;
    cin >> pi;
    if (pi > i) {
      tree.update(0, pi - i, pi - i, -1);
      tree.update(pi - i, n - i + 1, 0, 1);
      tree.update(n - i + 1, n, pi - 1, -1);
    } else if (pi < i) {
      tree.update(0, n - i + 1, i - pi, 1);
      tree.update(n - i + 1, n - i + pi, pi - 1, -1);
      tree.update(n - i + pi, n, 0, 1);
    } else {
      tree.update(1, n - i + 1, 1, 1);
      tree.update(n - i + 1, n, pi - 1, -1);
    }
  }
  pair<ll, int> best(1e18, -1);
  for (int k = 0; k < n; k++) {
    best = min(best, {tree.query(k), k});
  }
  cout << best.first << ' ' << best.second << '\n';
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using PII = pair<int, int>;
using VI = vector<int>;
class SegmentTree {
  int sz;
  vector<ll> add, dif;

 public:
  SegmentTree(int n) {
    for (sz = 1; sz < n; sz <<= 1)
      ;
    add.resize(2 * sz);
    dif.resize(2 * sz);
  }
  void update(int l, int r, ll a, ll d, int node = 1, int x = 0, int y = -1) {
    if (y == -1) y = sz;
    if (l >= y || r <= x) return;
    if (l <= x && y <= r) {
      add[node] += a + (x - l) * d;
      dif[node] += d;
      return;
    }
    update(l, r, a, d, 2 * node, x, (x + y) / 2);
    update(l, r, a, d, 2 * node + 1, (x + y) / 2, y);
  }
  ll query(int i, int node = 1, int x = 0, int y = -1) {
    if (y == -1) y = sz;
    if (i >= y || i < x) return 0;
    if (y - x == 1) return add[node];
    ll res = add[node] + (i - x) * dif[node];
    int z = (x + y) / 2;
    if (i < z)
      res += query(i, 2 * node, x, z);
    else
      res += query(i, 2 * node + 1, z, y);
    return res;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  SegmentTree tree(n + 1);
  for (int i = 1; i <= n; i++) {
    int pi;
    cin >> pi;
    if (pi > i) {
      tree.update(0, pi - i, pi - i, -1);
      tree.update(pi - i, n - i + 1, 0, 1);
      tree.update(n - i + 1, n, pi - 1, -1);
    } else if (pi < i) {
      tree.update(0, n - i + 1, i - pi, 1);
      tree.update(n - i + 1, n - i + pi, pi - 1, -1);
      tree.update(n - i + pi, n, 0, 1);
    } else {
      tree.update(1, n - i + 1, 1, 1);
      tree.update(n - i + 1, n, pi - 1, -1);
    }
  }
  pair<ll, int> best(1e18, -1);
  for (int k = 0; k < n; k++) {
    best = min(best, {tree.query(k), k});
  }
  cout << best.first << ' ' << best.second << '\n';
  return 0;
}

```