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 = 1010;
const int maxv = 8;
const int INF = 1000000000;
vector<int> in[maxv];
int dp[maxn][1 << maxv], cur[maxn], A[maxn];
void getIn(int n) {
  for (int i = 0; i < maxv; i++) in[i].clear();
  for (int i = 0; i < n; i++) in[A[i]].push_back(i);
}
char ss[10];
void getSS(int j) {
  for (int i = 0; i < 8; i++) {
    if (j & (1 << i))
      ss[7 - i] = '1';
    else
      ss[7 - i] = '0';
  }
}
int solve(int n, int len) {
  memset(cur, 0, sizeof(cur));
  for (int i = 0; i <= n; i++)
    for (int j = 0; j < (1 << maxv); j++) dp[i][j] = -INF;
  dp[0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < (1 << maxv); j++) {
      if (dp[i][j] == -INF) continue;
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      if (j & (1 << A[i])) continue;
      int it = cur[A[i]] + len - 1;
      if (it < in[A[i]].size())
        dp[in[A[i]][it] + 1][j | (1 << A[i])] =
            max(dp[in[A[i]][it] + 1][j | (1 << A[i])], dp[i][j]);
      it++;
      if (it < in[A[i]].size()) {
        dp[in[A[i]][it] + 1][j | (1 << A[i])] =
            max(dp[in[A[i]][it] + 1][j | (1 << A[i])], dp[i][j] + 1);
      }
    }
    cur[A[i]]++;
  }
  int ans = -INF;
  for (int i = 0; i <= n; i++) ans = max(ans, dp[i][(1 << maxv) - 1]);
  return ans;
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    for (int i = 0; i < n; i++) scanf("%d", &A[i]), A[i]--;
    getIn(n);
    int ans = 0, x = 0, y = n / 8, las = 0;
    while (x <= y) {
      int m = (x + y) / 2;
      int res = solve(n, m);
      if (res != -INF) {
        ans = max(ans, res * (m + 1) + (8 - res) * m);
        x = m + 1;
        las = m;
      } else
        y = m - 1;
    }
    int tmp = 0;
    for (int i = 0; i < maxv; i++)
      if (!in[i].empty()) tmp++;
    printf("%d\n", max(ans, tmp));
  }
  return 0;
}
