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;
inline int read() {
  register int x = 0, f = 1;
  register char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c & 15);
  return x * f;
}
template <typename T>
inline bool cmax(T& x, T y) {
  return x < y ? (x = y), 1 : 0;
}
template <typename T>
inline bool cmin(T& x, T y) {
  return x > y ? (x = y), 1 : 0;
}
template <typename T>
inline bool cabs(T& x) {
  return x > 0 ? 1 : (x = -x), 0;
}
inline int QP(int x, int y, int Mod) {
  int ans = 1;
  for (; y; y >>= 1, x = (x * x) % Mod)
    if (y & 1) ans = (ans * x) % Mod;
  return ans;
}
int n;
vector<int> v[12];
int dp[1 << 11][1 << 9];
int a[1 << 11];
int num[12];
int ans = 0;
inline int max(int x, int y) { return x > y ? x : y; }
inline bool chk(int mid) {
  memset(num, 0, sizeof(num));
  memset(dp, 250, sizeof(dp));
  dp[0][0] = 0;
  int tot = -0x3f3f3f3f;
  for (register int i = 0; i < n; i++) {
    for (register int j = 0; j <= (1 << 8) - 1; j++)
      if (dp[i][j] >= 0) {
        for (register int k = 0; k <= 7; k++) {
          if ((j & (1 << (k)))) continue;
          int now = num[k] + mid - 1;
          if (now < v[k].size())
            dp[v[k][now]][j | (1 << (k))] =
                max(dp[v[k][now]][j | (1 << (k))], dp[i][j]);
          if (++now < v[k].size())
            dp[v[k][now]][j | (1 << (k))] =
                max(dp[v[k][now]][j | (1 << (k))], dp[i][j] + 1);
        }
      }
    ++num[a[i]];
  }
  for (register int i = 0; i <= n; i++) cmax(tot, dp[i][(1 << 8) - 1]);
  if (tot < 0)
    return 0;
  else {
    cmax(ans, (mid << 3) + tot);
    return 1;
  }
}
signed main() {
  n = read();
  a[0] = 9;
  for (register int i = 1; i <= n; i++) {
    a[i] = read() - 1;
    v[a[i]].push_back(i);
  }
  int l = 0, r = n >> 3;
  int mn = n;
  for (register int i = 0; i <= 7; i++) mn = min(mn, (int)(v[i].size()));
  if (!mn) {
    for (register int i = 0; i <= 7; i++)
      if (v[i].size()) ans++;
    printf("%d", ans);
    return 0;
  }
  while (l <= r) {
    int mid = l + r >> 1;
    if (chk(mid)) {
      l = mid + 1;
    } else {
      r = mid - 1;
    }
  }
  printf("%d", ans);
  return 0;
}
