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 MOD = 1000000007;
const int N = 1000005;
const double PI = 4 * atan(1);
const double eps = 1e-7;
long long n;
long long dp[1005][256];
bool vis[1005][256];
long long a[1005];
long long troll[1005][1005];
long long t[12];
long long solve(long long pos, long long mask, long long toul) {
  if (pos >= n) {
    if (toul == 0) {
      return 0;
    } else {
      if (mask != (1 << 8) - 1) return -1e18;
      return 0;
    }
  }
  if (vis[pos][mask]) return dp[pos][mask];
  vis[pos][mask] = 1;
  long long to_return = solve(pos + 1, mask, toul);
  if (((1 << a[pos]) & mask) == 0) {
    long long x = troll[pos][toul];
    if (x != -1) {
      to_return =
          max(to_return, toul + solve(x + 1, mask + (1 << a[pos]), toul));
    }
    x = troll[pos][toul + 1];
    if (x != -1) {
      to_return =
          max(to_return, toul + 1 + solve(x + 1, mask + (1 << a[pos]), toul));
    }
  }
  return dp[pos][mask] = to_return;
}
long long possible(long long u) {
  memset(dp, -1, sizeof(dp));
  memset(vis, 0, sizeof(vis));
  long long ans = solve(0, 0, u);
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  memset(dp, -1, sizeof(dp));
  memset(troll, -1, sizeof(troll));
  cin >> n;
  long long ma = 10000;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
    t[a[i]]++;
  }
  for (int i = 0; i < n; i++) {
    long long cnt = 0;
    for (int j = i; j < n; j++) {
      if (a[j] == a[i]) {
        cnt++;
        troll[i][cnt] = j;
      }
    }
  }
  for (int i = 0; i < 8; i++) {
    ma = min(ma, t[i]);
  }
  long long l = 0;
  long long r = ma;
  while (l < r - 1) {
    long long mid = (l + r) / 2;
    if (possible(mid) >= 0) {
      l = mid;
    } else {
      r = mid - 1;
    }
  }
  long long x = possible(r);
  if (x >= 0) {
    cout << x;
  } else {
    x = possible(l);
    cout << x << endl;
  }
  return 0;
}
