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;
int n;
int A[1016];
int G[1016][9][1016], lsj[9];
bool chk(int x, int s) {
  for (int i = (1), iend = (8); i <= iend; ++i) lsj[i] = i;
  int cur = 0;
  do {
    cur = 0;
    for (int k = (1), kend = (8); k <= kend; ++k) {
      cur = G[cur][lsj[k]][x + ((s >> lsj[k] - 1) & 1)];
      if (cur == -1) break;
    }
    if (~cur) return true;
  } while (next_permutation(lsj + 1, lsj + 1 + 8));
  return false;
}
void solve() {
  cin >> n;
  for (int i = (1), iend = (n); i <= iend; ++i) scanf("%1d", A + i);
  memset(G, -1, sizeof G);
  for (int i = (n), iend = (1); i >= iend; --i) {
    for (int k = (1), kend = (8); k <= kend; ++k) G[i - 1][k][1] = G[i][k][1];
    G[i - 1][A[i]][1] = i;
    for (int k = (1), kend = (8); k <= kend; ++k)
      if (~G[i - 1][k][1]) {
        for (int j = (2), jend = (n); j <= jend; ++j)
          if (~G[G[i - 1][k][j - 1]][k][1]) {
            G[i - 1][k][j] = G[G[i - 1][k][j - 1]][k][1];
          } else
            break;
      }
  }
  int l = 1, r = n / 8;
  while (l <= r) {
    int mid = l + r >> 1;
    if (chk(mid, 0))
      l = mid + 1;
    else
      r = mid - 1;
  }
  if (r == 0) {
    int as = 0;
    for (int k = (1), kend = (8); k <= kend; ++k)
      if (~G[0][k][1]) ++as;
    printf("%d\n", as);
    return;
  }
  int ans = r * 8;
  for (int s = 1; s < (1 << 8); ++s) {
    if (chk(r, s)) ans = max(ans, __builtin_popcount(s) + r * 8);
  }
  printf("%d\n", ans);
}
signed main() { solve(); }
