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 long long mod = 1e9 + 7;
int n, a[1003];
vector<int> vec[8];
int dpL[(1 << 8)][1003], dpL1[(1 << 8)][1003];
int getVal(int mask, int pos, int len) {
  return dpL[mask][pos] * len + dpL1[mask][pos] * (len + 1);
}
void work(int len, int fst, int mask, int oldMask) {
  int bitSet = 0;
  for (int i = 0; i < 8; i++)
    if (oldMask & (1 << i)) bitSet++;
  if (vec[fst].size() < len) return;
  if (len == 0)
    for (int i = 1; i <= n; i++)
      if (getVal(mask, i, len) < getVal(oldMask, i, len) &&
          dpL[oldMask][i] + dpL1[oldMask][i] == bitSet)
        dpL[mask][i] = dpL[oldMask][i] + 1, dpL1[mask][i] = dpL1[oldMask][i];
  for (int i = 0; i <= vec[fst].size() - len; i++) {
    if (len == 0) break;
    int pos = vec[fst][i] - 1, l = vec[fst][i + len - 1],
        r = (i == vec[fst].size() - len ? n : vec[fst][i + len] - 1);
    for (int j = l; j <= r; j++) {
      if (getVal(mask, j, len) < getVal(oldMask, pos, len) + len &&
          dpL[oldMask][pos] + dpL1[oldMask][pos] == bitSet) {
        dpL[mask][j] = dpL[oldMask][pos] + 1;
        dpL1[mask][j] = dpL1[oldMask][pos];
      }
    }
  }
  if (vec[fst].size() == len) return;
  for (int i = 0; i < vec[fst].size() - len; i++) {
    int pos = vec[fst][i] - 1, l = vec[fst][i + len],
        r = (i == vec[fst].size() - len - 1 ? n : vec[fst][i + len + 1] - 1);
    for (int j = l; j <= r; j++) {
      if (getVal(mask, j, len) < getVal(oldMask, pos, len) + len + 1 &&
          dpL[oldMask][pos] + dpL1[oldMask][pos] == bitSet) {
        dpL[mask][j] = dpL[oldMask][pos];
        dpL1[mask][j] = dpL1[oldMask][pos] + 1;
      }
    }
  }
  return;
}
int calc(int len) {
  memset(dpL, 0, sizeof dpL);
  memset(dpL1, 0, sizeof dpL1);
  for (int mask = 1; mask < (1 << 8); mask++) {
    for (int fst = 0; fst < 8; fst++) {
      if ((mask & (1 << fst)) == 0) continue;
      work(len, fst, mask, mask - (1 << fst));
    }
  }
  return getVal((1 << 8) - 1, n, len);
}
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) a[i]--;
  for (int i = 1; i <= n; i++) vec[a[i]].push_back(i);
  int l = 0, r = n;
  while (1) {
    if (l == r) break;
    if (l == r - 1) {
      if (calc(r) >= r * 8 && dpL[(1 << 8) - 1][n] + dpL1[(1 << 8) - 1][n] == 8)
        l = r;
      else
        r = l;
      break;
    }
    if (calc((l + r >> 1)) >= (l + r >> 1) * 8 &&
        dpL[(1 << 8) - 1][n] + dpL1[(1 << 8) - 1][n] == 8)
      l = (l + r >> 1);
    else
      r = (l + r >> 1);
  }
  cout << calc(l);
  return;
}
int main() {
  solve();
  return 0;
}
