Vladik was bored on his way home and decided to play the following game. He took n cards and put them in a row in front of himself. Every card has a positive integer number not exceeding 8 written on it. He decided to find the longest subsequence of cards which satisfies the following conditions:

  * the number of occurrences of each number from 1 to 8 in the subsequence doesn't differ by more then 1 from the number of occurrences of any other number. Formally, if there are ck cards with number k on them in the subsequence, than for all pairs of integers <image> the condition |ci - cj| ≤ 1 must hold. 
  * if there is at least one card with number x on it in the subsequence, then all cards with number x in this subsequence must form a continuous segment in it (but not necessarily a continuous segment in the original sequence). For example, the subsequence [1, 1, 2, 2] satisfies this condition while the subsequence [1, 2, 2, 1] doesn't. Note that [1, 1, 2, 2] doesn't satisfy the first condition. 



Please help Vladik to find the length of the longest subsequence that satisfies both conditions.

Input

The first line contains single integer n (1 ≤ n ≤ 1000) — the number of cards in Vladik's sequence.

The second line contains the sequence of n positive integers not exceeding 8 — the description of Vladik's sequence.

Output

Print single integer — the length of the longest subsequence of Vladik's sequence that satisfies both conditions.

Examples

Input

3
1 1 1


Output

1

Input

8
8 7 6 5 4 3 2 1


Output

8

Input

24
1 8 1 2 8 2 3 8 3 4 8 4 5 8 5 6 8 6 7 8 7 8 8 8


Output

17

Note

In the first sample all the numbers written on the cards are equal, so you can't take more than one card, otherwise you'll violate the first condition.
#include <bits/stdc++.h>
using namespace std;
int n, l, r;
vector<int> a;
int b[8], su;
vector<int> ne[8];
vector<int> xe;
bool ok(int x) {
  int v[8];
  for (int i = 0; i < 8; ++i) v[i] = i;
  do {
    int k = 0;
    for (int i = 0; k < n && i < 8; ++i) {
      if (a[k] != v[i]) k = ne[v[i]][k];
      for (int j = 0; j < x - 1; ++j) k = ne[v[i]][k];
    }
    if (k < n) return true;
  } while (next_permutation(v, v + 8));
  return false;
}
bool okk() {
  int v[8];
  for (int i = 0; i < 8; ++i) v[i] = i;
  do {
    int k = 0;
    for (int i = 0; k < n && i < 8; ++i) {
      if (l == 0) {
        if (b[i] == 1) {
          if (a[k] != v[i]) k = ne[v[i]][k];
        }
      } else {
        if (a[k] != v[i]) k = ne[v[i]][k];
        k = xe[k];
        if (b[i] == l + 1) k = ne[v[i]][k];
      }
    }
    if (k < n) return true;
  } while (next_permutation(v, v + 8));
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  a.resize(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    --a[i];
  }
  for (int i = 0; i < 8; ++i) ne[i].assign(n + 1, n);
  for (int i = 0; i < 8; ++i) {
    for (int j = 0; j < n; ++j) {
      int k = j + 1;
      while (k < n && a[k] != i) ++k;
      ne[i][j] = k;
    }
  }
  l = 0;
  r = 150;
  while (r - l > 1) {
    int mid = (r + l) / 2;
    (ok(mid) ? l : r) = mid;
  }
  xe.assign(n + 1, n);
  for (int i = 0; i < n; ++i) {
    int j = i;
    for (int h = 0; h < l - 1; ++h) j = ne[a[i]][j];
    xe[i] = j;
  }
  int res = l * 8;
  int lim = (1 << 8);
  for (int i = 0; i < lim; ++i) {
    int x = i;
    su = 0;
    for (int j = 0; j < 8; ++j) {
      b[j] = l + x % 2;
      x /= 2;
      su += b[j];
    }
    if (res < su) {
      if (okk()) res = su;
    }
  }
  cout << res << endl;
  return 0;
}
