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 MOD = 1e9 + 7;
const int MAX = 1e3 + 5;
int f[9], n, freq;
int bit[9][MAX], dp[(1 << 9)][MAX];
int _find(int id, int i, int ff) {
  int lo = i, hi = n, pl = -1, ph = -1;
  while (1) {
    int mid = (lo + hi) >> 1;
    if (pl == lo && ph == hi) break;
    pl = lo;
    ph = hi;
    if (bit[id][mid] - bit[id][i - 1] >= ff)
      hi = mid;
    else
      lo = mid + 1;
  }
  return lo;
}
int solve(int mask, int i) {
  if (i == n + 1) {
    if (mask != (1 << 8) - 1) return -1e9;
    return 0;
  }
  if (dp[mask][i] != -1) return dp[mask][i];
  int ret = solve(mask, i + 1);
  for (int j = 0; j < 8; j++) {
    if (mask & (1 << j)) continue;
    int x = _find(j + 1, i, freq);
    int y = _find(j + 1, i, freq + 1);
    if (freq == 0)
      ret = max(ret, solve(mask | (1 << j), i));
    else if (x <= n && bit[j + 1][x] - bit[j + 1][i - 1] == freq)
      ret = max(ret, freq + solve(mask | (1 << j), x + 1));
    if (y <= n && bit[j + 1][y] - bit[j + 1][i - 1] == freq + 1)
      ret = max(ret, freq + 1 + solve(mask | (1 << j), y + 1));
  }
  return dp[mask][i] = ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    f[x]++;
    bit[x][i] = 1;
  }
  for (int i = 1; i <= 8; i++) {
    for (int j = 1; j < MAX; j++) {
      bit[i][j] += bit[i][j - 1];
    }
  }
  int ret = 1;
  int lo = 0, hi = n, pl = -1, ph = -1;
  while (1) {
    freq = lo + hi >> 1;
    if (pl == lo && ph == hi) break;
    pl = lo;
    ph = hi;
    memset(dp, -1, sizeof dp);
    int t = solve(0, 1);
    if (t > 0) {
      lo = freq;
      ret = max(ret, t);
    } else
      hi = freq;
  }
  for (freq = lo; freq <= hi; freq++) ret = max(ret, solve(0, 1));
  cout << ret;
  return 0;
}
