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>
#pragma GCC optimize("Ofast, unroll-loops", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native,avx2")
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int getrnd(int l, int r) { return uniform_int_distribution<int>(l, r)(rng); }
template <typename T1, typename T2>
bool relax(T1& a, const T2& b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T1, typename T2>
bool strain(T1& a, const T2& b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
const int N = 1005, C = 8;
int n, a[N], dp[N][1 << C];
int good(int len) {
  if (len == 0) {
    vector<int> cnt(8, 0);
    for (int i = 0; i < n; ++i) ++cnt[a[i]];
    int x = 0;
    for (int i = 0; i < 8; ++i)
      if (cnt[i] > 0) ++x;
    return x;
  }
  for (int i = 0; i <= n; ++i)
    for (int msk = 0; msk < (1 << C); ++msk) dp[i][msk] = -1;
  dp[0][0] = 0;
  for (int i = 0; i < n; ++i) {
    int nmx = 0;
    for (int msk = 0; msk < (1 << C); ++msk) {
      strain(dp[i + 1][msk], dp[i][msk]);
      if (dp[i][msk] < 0) continue;
      if (!((msk >> a[i]) & 1)) {
        int c = 0, nmsk = msk | (1 << a[i]);
        for (int j = i; j < n; ++j) {
          if (a[j] == a[i]) ++c;
          if (c == len) {
            strain(dp[j + 1][nmsk], dp[i][msk]);
          } else if (c >= len + 1) {
            strain(dp[j + 1][nmsk], dp[i][msk] + 1);
          }
        }
      }
    }
  }
  return dp[n][(1 << C) - 1];
}
void solve() {
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i], --a[i];
  int l = 0, r = n + 1, mid;
  while (r - l > 1) {
    mid = (r + l) / 2;
    if (good(mid) >= 0)
      l = mid;
    else
      r = mid;
  }
  cout << 8 * l + good(l) << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  srand(time(0));
  int t = 1;
  while (t--) solve();
  return 0;
}
