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 = 1007;
const int inf = 1000000000;
int f[maxn][1 << 8][200], n, luu[maxn], res, RES, nx[maxn][10], a[maxn];
int dp(int x, int y, int z) {
  if (x == n + 1) {
    if (z < RES) return -inf;
    if (y != (1 << 8) - 1) return -inf;
    return 0;
  }
  if (f[x][y][z] == -1) {
    f[x][y][z] = -inf;
    if (z <= RES)
      f[x][y][z] = dp(nx[x][a[x]], y, z + (a[nx[x][a[x]]] == a[x])) + 1;
    if (z == RES || z == RES + 1)
      for (int i = 1; i <= 8; ++i)
        if (i != a[x]) {
          if (y & (1 << (i - 1))) continue;
          f[x][y][z] =
              max(f[x][y][z],
                  dp(nx[x][i], y | (1 << (i - 1)), a[nx[x][i]] == i) + 1);
        }
    f[x][y][z] = max(f[x][y][z], dp(n + 1, y, z) + 1);
  }
  return f[x][y][z];
}
void solvedp(int rr) {
  memset(f, -1, sizeof f);
  RES = rr;
  for (int i = 1; i <= 8; ++i)
    f[0][0][0] = max(f[0][0][0], dp(luu[i], 1 << (i - 1), 1));
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int i = 1; i <= 8; ++i) luu[i] = n + 1;
  for (int i = n; i >= 1; --i) {
    for (int j = 1; j <= 8; ++j) nx[i][j] = luu[j];
    luu[a[i]] = i;
  }
  for (int i = 1; i <= 8; ++i)
    if (luu[i] != n + 1) res++;
  int l = 1, r = n / 8;
  while (l <= r) {
    int mid = (l + r) >> 1;
    solvedp(mid);
    if (f[0][0][0] < 0)
      r = mid - 1;
    else
      res = max(res, f[0][0][0]), l = mid + 1;
  }
  cout << res;
}
