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 v[1100];
int pd[1100][1 << 8];
int qtd1, qtd2;
int vis[1100][1 << 8];
int inv[1100];
vector<int> pos[10];
bool check(int mask) {
  for (int i = 0; i < 8; ++i) {
    if (mask & (1 << i)) continue;
    if (qtd1) return false;
  }
  return true;
}
int f(int x, int mask) {
  if (x >= n) {
    if (check(mask))
      return 0;
    else
      return -0x3F3F3F3F;
  }
  int &p = pd[x][mask];
  if (vis[x][mask] < qtd2) {
    vis[x][mask] = qtd2;
    p = f(x + 1, mask);
    if (!(mask & (1 << v[x]))) {
      int fim, qtd;
      qtd = pos[v[x]].size() - inv[x];
      if (qtd1 && qtd >= qtd1) {
        fim = pos[v[x]][inv[x] + qtd1 - 1] + 1;
        p = max(p, f(fim, mask | (1 << v[x])) + qtd1);
      }
      qtd = pos[v[x]].size() - inv[x];
      if (qtd >= qtd2) {
        fim = pos[v[x]][inv[x] + qtd2 - 1] + 1;
        p = max(p, f(fim, mask | (1 << v[x])) + qtd2);
      }
    }
  }
  return p;
}
int main() {
  int ans = 1;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &v[i]);
    v[i]--;
    inv[i] = (int)pos[v[i]].size();
    pos[v[i]].push_back(i);
  }
  for (int i = 0; i < n; ++i) {
    qtd1 = i, qtd2 = i + 1;
    ans = max(ans, f(0, 0));
  }
  printf("%d\n", ans);
  return 0;
}
