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, zong, ans;
int num[1003];
int jiyi[1003][1003];
int cnt[1003], nxt[1003][10][520];
int dp(int x, int S) {
  int &fanhui = jiyi[x][S];
  if (fanhui != -1) {
    return fanhui;
  }
  if (S == ((1 << 8) - 1)) {
    return 0;
  }
  if (x == n + 1) {
    return -99999999;
  }
  fanhui = -99999999;
  for (int i = 1; i <= 8; i++) {
    if (!(S & (1 << (i - 1)))) {
      for (int j = zong - 1; j <= zong; j++) {
        if (nxt[x][i][j] != 100000000) {
          fanhui = max(fanhui, dp(nxt[x][i][j] + 1, S ^ (1 << (i - 1))) + j);
        }
      }
    }
  }
  return fanhui;
}
inline void solve(int x) {
  zong = x;
  memset(jiyi, -1, sizeof(jiyi));
  ans = -99999999;
  for (int i = 1; i <= 8; i++) {
    for (int j = zong - 1; j <= zong; j++) {
      if (nxt[1][i][j] != 100000000) {
        ans = max(ans, dp(nxt[1][i][j] + 1, (1 << (i - 1))) + j);
      }
    }
  }
}
int tong[13], tot;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> num[i];
    if (!tong[num[i]]) {
      tong[num[i]] = 1;
      tot++;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= 8; j++) {
      cnt[j] = 0;
    }
    for (int j = 1; j <= 8; j++) {
      for (int k = 1; k <= n; k++) {
        nxt[i][j][k] = 1e8;
      }
    }
    for (int j = i; j <= n; j++) {
      cnt[num[j]]++;
      nxt[i][num[j]][cnt[num[j]]] = j;
    }
  }
  int l = 2, r = n;
  while (l < r - 1) {
    int mid = (l + r) / 2;
    solve(mid);
    if (ans > 0) {
      l = mid;
    } else {
      r = mid - 1;
    }
  }
  solve(r);
  if (ans > 0) {
    cout << max(tot, ans) << endl;
  } else {
    solve(l);
    cout << max(tot, ans) << endl;
  }
  return 0;
}
