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 N = 1e3 + 100;
const int ninf = 0xf7f7f7f7;
int a[N], n;
int f[1 << 8][N], pos[N];
vector<int> v[8];
inline int nxt(int x, int k) {
  int sz = v[a[x]].size();
  if (pos[x] + k >= sz)
    return -1;
  else
    return v[a[x]][pos[x] + k];
}
void init() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i), --a[i];
  for (int i = 1; i <= n; ++i) {
    v[a[i]].push_back(i);
    pos[i] = v[a[i]].size() - 1;
  }
}
inline void upd(int &x, int y) { x = max(x, y); }
int main() {
  init();
  int ans = 0;
  for (int i = 0; i <= 7; ++i) ans += v[i].size() > 0;
  for (int i = 0; i <= n / 8; ++i) {
    memset(f, 0xf7, sizeof f);
    f[0][0] = 0;
    for (int j = 0; j <= n - 1; ++j) {
      for (int k = 0; k <= 255; ++k) {
        if (f[k][j] == ninf) continue;
        f[k][j + 1] = max(f[k][j + 1], f[k][j]);
        if ((k >> a[j + 1]) & 1) continue;
        int z0 = nxt(j + 1, i), z1 = nxt(j + 1, i + 1);
        if (z0 != -1) {
          upd(f[k | (1 << a[j + 1])][z0], f[k][j] + i + 1);
        }
        if (z1 != -1) {
          upd(f[k | (1 << a[j + 1])][z1], f[k][j] + i + 2);
        }
      }
    }
    ans = max(ans, f[255][n]);
  }
  printf("%d\n", ans);
  return 0;
}
