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 INF = (1 << 30);
const int maxn = 1010;
int n, vis[10], ans, a[maxn], nex[maxn][maxn], dp[maxn][maxn];
void init() {
  memset(vis, 0, sizeof(vis));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    a[i]--;
  }
  for (int i = 1; i <= n; i++) {
    int k = 1, j = i;
    nex[i][k++] = i;
    if (vis[a[i]] == 0) {
      ans++;
      vis[a[i]] = 1;
    }
    while (++j <= n) {
      if (a[j] == a[i]) nex[i][k++] = j;
    }
  }
}
int dfs(int pos, int s, int l) {
  if (dp[pos][s] != -1) return dp[pos][s];
  if (s == 0) return dp[pos][s] = 0;
  int nn = -INF;
  if (pos > n) return dp[pos][s] = nn;
  nn = max(nn, dfs(pos + 1, s, l));
  if ((1 << a[pos]) & s) {
    if (nex[pos][l])
      nn = max(nn, dfs(nex[pos][l] + 1, s ^ (1 << a[pos]), l) + l);
    if (nex[pos][l + 1])
      nn = max(nn, dfs(nex[pos][l + 1] + 1, s ^ (1 << a[pos]), l) + l + 1);
  }
  return dp[pos][s] = nn;
}
bool check(int l) {
  memset(dp, -1, sizeof(dp));
  dfs(1, (1 << 8) - 1, l);
  return dp[1][(1 << 8) - 1] > 0;
}
void sov() {
  int left = 0, right = n;
  while (left < right) {
    int mid = (left + right + 1) >> 1;
    if (check(mid)) {
      left = mid;
      ans = max(ans, dp[1][(1 << 8) - 1]);
    } else
      right = mid - 1;
  }
  printf("%d\n", ans);
}
int main() {
  init();
  sov();
}
