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;
template <typename T>
void sci(T& t) {
  cin >> t;
}
template <typename T, typename... Ts>
void sci(T& t, Ts&... ts) {
  sci(t);
  sci(ts...);
}
int a[1111];
int d[1111][333];
int rt[1111][1111];
int nxt[1111];
int lst[9];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  sci(n);
  int mxk = (n + 7) / 8 + 31;
  mxk = min(mxk, n);
  memset(rt, -1, sizeof(int) * 1111 * 1111);
  memset(nxt, -1, sizeof(nxt));
  memset(lst, -1, sizeof(lst));
  for (int i = 0; i < n; i++) {
    sci(a[i]);
    rt[i][1] = i + 1;
  }
  for (int i = n - 1; i >= 0; i--) {
    int x = a[i];
    if (lst[x] != -1) {
      nxt[i] = lst[x];
    }
    lst[x] = i;
  }
  int ans = 0;
  for (int i = 0; i < 9; i++) {
    if (lst[i] != -1) {
      ans++;
    }
  }
  for (int k = 2; k <= mxk; k++) {
    for (int i = 0; i < n; i++) {
      if (nxt[i] != -1) {
        rt[i][k] = rt[nxt[i]][k - 1];
      }
    }
  }
  for (int k = 2; k <= mxk; k++) {
    memset(d, -1, sizeof(int) * 1111 * 333);
    d[0][0] = 0;
    for (int i = 0; i < n; i++) {
      for (int msk = 0; msk < (1 << 8); msk++) {
        if (d[i][msk] == -1) {
          continue;
        }
        d[i + 1][msk] = max(d[i + 1][msk], d[i][msk]);
        int x = a[i] - 1;
        if ((msk >> x) & 1) {
          continue;
        }
        if (rt[i][k] != -1) {
          d[rt[i][k]][msk | (1 << x)] =
              max(d[rt[i][k]][msk | (1 << x)], d[i][msk] + k);
        }
        if (rt[i][k - 1] != -1) {
          d[rt[i][k - 1]][msk | (1 << x)] =
              max(d[rt[i][k - 1]][msk | (1 << x)], d[i][msk] + k - 1);
        }
      }
    }
    ans = max(ans, d[n][(1 << 8) - 1]);
  }
  cout << ans;
  return 0;
}
