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;
int dp[(1 << 8) + 10][1000 + 10];
int mx[(1 << 8) + 10][1000 + 10];
int nx[8 + 2][125 + 2][1000 + 10];
int A[1000 + 10];
const int INF = 1e9;
void init() {
  memset(nx, -1, sizeof(nx));
  for (int u = 1; u <= 8; u++) {
    nx[u][0][N + 1] = N + 1;
    for (int i = N; i >= 1; i--) {
      nx[u][0][i] = i;
      for (int b = 1; b <= (N / 8 + 1); b++) {
        if (A[i] == u)
          nx[u][b][i] = nx[u][b - 1][i + 1];
        else
          nx[u][b][i] = nx[u][b][i + 1];
      }
    }
  }
}
int sset(int st, int x) { return st | (1 << (x - 1)); }
bool is(int st, int x) { return st & (1 << (x - 1)); }
int solve(int block) {
  if (block == 0) {
    int res = 0;
    for (int x = 1; x <= 8; x++) {
      res += count(A + 1, A + N + 1, x) > 0;
    }
    return res;
  }
  memset(dp, 0, sizeof(dp));
  memset(mx, 0, sizeof(mx));
  for (int st = 0; st < (1 << 8) - 1; st++) {
    dp[st][N + 1] = -INF;
    mx[st][N + 1] = -INF;
  }
  for (int i = 1; i <= N + 1; i++) {
    dp[(1 << 8) - 1][i] = 0;
    mx[(1 << 8) - 1][i] = 0;
  }
  for (int st = (1 << 8) - 2; st >= 0; st--) {
    for (int i = N; i >= 1; i--) {
      dp[st][i] = -INF;
      for (int x = 1; x <= 8; x++) {
        if (is(st, x)) continue;
        int l = nx[x][block][i];
        int r = nx[x][block + 1][i];
        if (l != -1 && mx[sset(st, x)][l] != -INF) {
          dp[st][i] = max(dp[st][i], block + mx[sset(st, x)][l]);
        }
        if (r != -1 && mx[sset(st, x)][r] != -INF) {
          dp[st][i] = max(dp[st][i], block + 1 + mx[sset(st, x)][r]);
        }
        mx[st][i] = max(dp[st][i], mx[st][i + 1]);
      }
    }
  }
  return dp[0][1];
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    scanf("%d", &A[i]);
  }
  init();
  int mx = 0;
  for (int d = 0; d <= N / 8; d++) {
    mx = max(mx, solve(d));
  }
  printf("%d\n", mx);
}
