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;
const int maxn = 1007;
const int sum = 1 << 8;
bool dp[maxn][sum][9];
bool dp2[maxn][sum];
int a[maxn], n;
int pre[maxn][9];
int pre2[maxn][9];
int cnt[9];
bool is_ok(int len) {
  if (len == 0) return true;
  int i, j, k, x, y;
  dp2[0][0] = 1;
  for (i = 1; i <= n; i++) {
    dp2[i][0] = 1;
    for (j = 1; j < sum; j++) {
      if (dp2[i - 1][j]) {
        dp2[i][j] = 1;
        continue;
      }
      dp2[i][j] = 0;
      for (k = 0; k < 8; k++) {
        if ((j & (1 << k)) > 0) {
          x = j - (1 << k);
          y = pre[i][k + 1];
          if (y >= len) {
            y = pre2[y - len + 1][k + 1] - 1;
            if (y >= 0 && dp2[y][x]) {
              dp2[i][j] = 1;
              break;
            }
          }
        }
      }
    }
  }
  return dp2[n][sum - 1];
}
int get_ans(int len) {
  int i, j, k, x, y, z;
  for (i = 0; i <= n; i++)
    for (j = 0; j < sum; j++)
      for (k = 0; k <= 8; k++) dp[i][j][k] = false;
  dp[0][0][0] = 1;
  for (i = 1; i <= n; i++) {
    dp[i][0][0] = 1;
    for (j = 1; j < sum; j++) {
      x = j;
      k = 0;
      while (x > 0) {
        if (x & 1) k++;
        x >>= 1;
      }
      for (; k >= 0; k--) {
        if (dp[i - 1][j][k]) {
          dp[i][j][k] = 1;
          continue;
        }
        for (x = 0; x < 8; x++) {
          if ((j & (1 << x)) > 0) {
            y = j - (1 << x);
            z = pre[i][x + 1];
            if (z >= len) {
              if (pre2[z - len + 1][x + 1] >= 1 &&
                  dp[pre2[z - len + 1][x + 1] - 1][y][k]) {
                printf("");
                dp[i][j][k] = 1;
                break;
              }
            }
            if (z > len) {
              if (pre2[z - len][x + 1] >= 1 &&
                  dp[pre2[z - len][x + 1] - 1][y][k - 1]) {
                printf("");
                dp[i][j][k] = 1;
                break;
              }
            }
          }
        }
      }
    }
  }
  for (i = 7; i >= 0; i--) {
    if (dp[n][sum - 1][i]) return i;
  }
}
int main() {
  int m, i, j, k, x, y;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  memset(pre2, -1, sizeof(pre2));
  for (i = 1; i <= n; i++) {
    pre[i][a[i]]++;
    cnt[a[i]]++;
    pre2[cnt[a[i]]][a[i]] = i;
    for (j = 1; j <= 8; j++) pre[i][j] += pre[i - 1][j];
  }
  int l, r, mid;
  l = 0, r = 1007;
  for (i = 1; i <= 8; i++) r = min(r, cnt[i]);
  r++;
  while (r - l > 1) {
    mid = (l + r) / 2;
    if (is_ok(mid))
      l = mid;
    else
      r = mid;
  }
  if (l == 0) {
    x = 0;
    for (i = 1; i <= 8; i++)
      if (cnt[i] > 0) x++;
    printf("%d\n", x);
  } else
    printf("%d\n", l * 8 + get_ans(l));
  return 0;
}
