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 cum[1011][8];
int A[1011];
int agla[1011][8];
int nx[1011][1011][8];
bool pre[10];
int n;
int ans = 0;
vector<int> perm = {1, 2, 3, 4, 5, 6, 7, 8};
inline bool canForm(int len, int mask) {
  int start = 1;
  int idx = 0;
  while (1) {
    start = nx[start][len + ((mask >> idx) & 1)][perm[idx]];
    if (start < 0) return false;
    if (idx == 7) return true;
    start++;
    idx++;
  }
}
inline void tryForm(int len) {
  if (len == 0) return;
  for (int mask = 0; mask < 256; mask++) {
    if (canForm(len, mask)) {
      ans = max(ans, ((len << 3) + __builtin_popcount(mask)));
    }
  }
}
int main() {
  memset(nx, -1, sizeof(nx));
  memset(agla, -1, sizeof(agla));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", A + i);
    pre[A[i]] = true;
    cum[i][A[i]] = 1;
    for (int j = 1; j <= 8; j++) {
      cum[i][j] += cum[i - 1][j];
    }
  }
  for (int i = n; i >= 1; i--) {
    for (int j = 1; j <= 8; j++) {
      if (A[i] == j)
        agla[i][j] = i;
      else
        agla[i][j] = agla[i + 1][j];
    }
  }
  for (int i = n; i >= 1; i--) {
    for (int j = 1; j <= n; j++) {
      assert(j != 0);
      for (int k = 1; k <= 8; k++) {
        if (j == 1)
          nx[i][j][k] = agla[i][k];
        else {
          int oneless = nx[i][j - 1][k];
          if (oneless == -1)
            nx[i][j][k] = -1;
          else
            nx[i][j][k] = agla[oneless + 1][k];
        }
      }
    }
  }
  for (int i = 1; i <= 8; i++) ans += pre[i];
  do {
    int l = 0, r = 150;
    while (l != r) {
      int mid = (l + r) >> 1;
      if (canForm(mid, 0))
        l = mid + 1;
      else
        r = mid;
    }
    tryForm(l - 1);
    tryForm(l - 2);
  } while (next_permutation(perm.begin(), perm.end()));
  printf("%d\n", ans);
}
