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 N = 1003;
const int M = (1 << 8);
int dp[N][M];
int n;
int g[N], cur[8];
vector<int> v[8];
int ok(int m) {
  memset((cur), (0), sizeof(cur));
  memset((dp), (-1), sizeof(dp));
  dp[0][0] = 0;
  for (int i = (int)0; i < (int)n; ++i) {
    for (int j = (int)0; j < (int)M; ++j)
      if (dp[i][j] != -1) {
        for (int k = (int)0; k < (int)8; ++k)
          if (!((1 << k) & j)) {
            int x = m + cur[k] - 1, y, z;
            if (x >= ((int)(v[k]).size())) continue;
            y = v[k][x];
            z = j | (1 << k);
            dp[y][z] = max(dp[y][z], dp[i][j]);
            if (++x >= ((int)(v[k]).size())) continue;
            y = v[k][x];
            dp[y][z] = max(dp[y][z], dp[i][j] + 1);
          }
      }
    ++cur[g[i]];
  }
  int x = -1;
  for (int i = (int)0; i < (int)n; ++i) x = max(x, dp[i][M - 1]);
  return x == -1 ? x : m * 8 + x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = (int)0; i < (int)n; ++i) {
    cin >> g[i];
    v[--g[i]].push_back(i);
  }
  int l = 1, r = (n >> 3) + 1, m;
  int res = 0, x;
  while (l + 1 <= r) {
    m = (l + r) >> 1;
    if ((x = ok(m)) != -1) {
      l = m + 1;
      res = max(res, x);
    } else
      r = m;
  }
  if (res == 0)
    for (int i = (int)0; i < (int)8; ++i)
      if (!v[i].empty()) ++res;
  cout << res;
  return 0;
}
