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 = 1e3 + 22;
int n, a[MAXN], p[9], t[9], res;
vector<int> v;
bool m[40322];
inline bool ck(int &x) {
  memset(t, 0, sizeof(t));
  int cur = 1;
  for (int i = 1; i <= n; i++) {
    if (a[i] == p[cur]) t[a[i]]++;
    if (t[p[cur]] == x) cur++;
  }
  return (cur == 9);
}
inline int f(int x) {
  int cnt = 0;
  while (x) {
    cnt += x % 2;
    x /= 2;
  }
  return cnt;
}
int count(int x) {
  int cnt = 0, mx = 0, cur;
  for (int i = 0; i < (1 << 8); i++) {
    memset(t, 0, sizeof(t));
    cur = 1;
    for (int j = 1; j <= n; j++) {
      while (cur <= 8 && t[p[cur]] == x + ((i & (1 << (p[cur] - 1))) != 0) &&
             cur <= 8)
        cur++;
      if (a[j] == p[cur]) t[a[j]]++;
    }
    while (cur <= 8 && t[p[cur]] == x + ((i & (1 << (p[cur] - 1))) != 0)) cur++;
    if (cur == 9) mx = max(mx, f(i));
  }
  return mx;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= 8; i++) p[i] = i;
  int cnt = 0;
  do {
    cnt++;
    int low = 1, high = 125, mid;
    while (low <= high) {
      mid = (low + high) / 2;
      if (ck(mid))
        low = mid + 1;
      else
        high = mid - 1;
    }
    if (res == high)
      v.push_back(cnt);
    else if (res < high) {
      v.clear();
      res = high;
      v.push_back(cnt);
    }
  } while (next_permutation(p + 1, p + 8 + 1));
  for (int i = 0; i < v.size(); i++) m[v[i]] = true;
  int t = 0;
  cnt = 0;
  do {
    cnt++;
    if (m[cnt]) t = max(t, count(res));
  } while (next_permutation(p + 1, p + 1 + 8));
  printf("%d", res * 8 + t);
  return 0;
}
