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>
#pragma comment(linker, "/STACK:32000000")
using namespace std;
inline bool read(int& val) { return scanf("%d", &val) != -1; }
inline bool read(long long& val) { return scanf("%I64d", &val) != -1; }
inline bool read(double& val) { return scanf("%lf", &val) != -1; }
inline bool read(char* val) { return scanf("%s", val) != -1; }
template <class T1, class T2>
inline bool read(T1& a, T2& b) {
  return read(a) && read(b);
}
template <class T1, class T2, class T3>
inline bool read(T1& a, T2& b, T3& c) {
  return read(a) && read(b) && read(c);
}
template <class T1, class T2, class T3, class T4>
inline bool read(T1& a, T2& b, T3& c, T4& d) {
  return read(a) && read(b) && read(c) && read(d);
}
template <class T1, class T2, class T3, class T4, class T5>
inline bool read(T1& a, T2& b, T3& c, T4& d, T5& e) {
  return read(a) && read(b) && read(c) && read(d) && read(e);
}
const int nil = -1;
const int N = 1111;
int dp[N][256];
int to[N][8][N];
int val[N];
int nbits[256];
int main() {
  for (int i = 0, _ = (256); i < _; ++i)
    for (int k = 0, _ = (8); k < _; ++k) nbits[i] += (bool)(i & (1 << k));
  int n;
  read(n);
  for (int i = 0, _ = (n); i < _; ++i) read(val[i]);
  for (int i = 0, _ = (n); i < _; ++i) val[i]--;
  for (int k = 0, _ = (8); k < _; ++k)
    for (int i = 0, _ = (n); i < _; ++i)
      for (int j = 0, _ = (n); j < _; ++j) to[i][k][j] = N;
  for (int k = 0, _ = (8); k < _; ++k)
    for (int i = 0, _ = (n); i < _; ++i) {
      int cnt = 0;
      to[0][k][i] = i;
      for (int j = (i), _ = (n); j < _; ++j)
        if (val[j] == k) to[++cnt][k][i] = j;
    }
  for (int i = 0, _ = (n + 1); i < _; ++i)
    for (int j = 0, _ = (258); j < _; ++j) dp[i][j] = N;
  for (int i = (1), _ = (n + 1); i < _; ++i)
    for (int k = 0, _ = (8); k < _; ++k) {
      dp[i][1 << k] = to[i][k][0];
    }
  for (int i = (1), _ = (n + 1); i < _; ++i)
    for (int j = 0, _ = (256); j < _; ++j)
      if (dp[i][j] != N) {
        int cnt = i / nbits[j];
        int one = i % nbits[j];
        for (int k = 0, _ = (8); k < _; ++k)
          if (!(j & (1 << k))) {
            int p = dp[i][j], s = j | (1 << k);
            if (i + cnt <= n) {
              int& nxt = dp[i + cnt][s];
              nxt = min(nxt, to[cnt][k][p]);
            }
            if (i + cnt + 1 <= n) {
              int& nxt = dp[i + cnt + 1][s];
              nxt = min(nxt, to[cnt + 1][k][p]);
            }
            if (i + cnt - 1 <= n && one == 0) {
              int& nxt = dp[i + cnt - 1][s];
              nxt = min(nxt, to[cnt - 1][k][p]);
            }
          }
      }
  int ans = 0;
  for (int i = 0, _ = (n + 1); i < _; ++i)
    if (dp[i][255] != N) ans = max(ans, i);
  printf("%d\n", ans);
  return 0;
}
