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;
bool dp[1005][256][8];
int a[1005], step[1005][1005], last[1005];
int main() {
  int N;
  scanf("%d", &N);
  memset(last, -1, sizeof last);
  memset(step, -1, sizeof step);
  for (int i = 1; i <= N; i++) {
    scanf("%d", a + i);
    int cur = 1;
    step[i][0] = i;
    step[i][1] = last[a[i]];
    last[a[i]] = i;
    while (step[i][cur] != -1) {
      step[i][cur + 1] = step[step[i][cur]][1];
      cur++;
    }
  }
  bool flag = true;
  for (int len = N; len >= 8; len--) {
    int t = len / 8, ex = len % 8;
    memset(dp, 0, sizeof dp);
    dp[0][0][0] = true;
    for (int i = 1; i <= N; i++) {
      memcpy(dp[i], dp[i - 1], sizeof dp[i]);
      for (int s = 1; s < 256; s++) {
        if (s & (1 << (a[i] - 1))) {
          for (int j = 0; j <= ex; j++) {
            if (step[i][t - 1] != -1)
              dp[i][s][j] |= dp[step[i][t - 1] - 1][s ^ (1 << a[i] - 1)][j];
            if (step[i][t] != -1 && j > 0)
              dp[i][s][j] |= dp[step[i][t] - 1][s ^ (1 << a[i] - 1)][j - 1];
          }
        }
      }
    }
    if (dp[N][255][ex]) {
      printf("%d\n", len);
      flag = false;
      break;
    }
  }
  if (flag) {
    for (int len = 7; len >= 1; len--) {
      bool vis[9] = {0};
      int cnt = 0;
      for (int i = 1; i <= N; i++) {
        if (!vis[a[i]]) {
          vis[a[i]] = true;
          cnt++;
        }
      }
      if (cnt >= len) {
        printf("%d\n", len);
        break;
      }
    }
  }
  return 0;
}
