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;
inline bool checkBit(long long n, int i) { return n & (1LL << i); }
inline long long setBit(long long n, int i) { return n | (1LL << i); }
inline long long resetBit(long long n, int i) { return n & (~(1LL << i)); }
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline double dist(double ix, double iy, double jx, double jy) {
  return sqrt(((ix - jx) * (ix - jx)) + ((iy - jy) * (iy - jy)));
}
int marked[1000000 / 64 + 2];
bool isPrime(int x) {
  return (x > 1) &&
         ((x == 2) ||
          ((x & 1) && (!((marked[x >> 6] & (1 << ((x & 63) >> 1)))))));
}
void seive(int n) {
  int i, j;
  for (i = 3; i * i <= n; i += 2) {
    if (!(marked[i >> 6] & (1 << ((i & 63) >> 1)))) {
      for (j = i * i; j <= n; j += i << 1) {
        marked[j >> 6] |= (1 << ((j & 63) >> 1));
      }
    }
  }
}
long long bigMod(long long a, long long b) {
  long long r = 1;
  while (b) {
    if (b & 1) (r *= a) %= 1000000007;
    b >>= 1;
    (a *= a) %= 1000000007;
  }
  return r;
}
long long add(long long a, long long b) {
  long long ret = a + b;
  if (ret >= 1000000007) ret -= 1000000007;
  return ret;
}
long long sub(long long a, long long b) {
  long long ret = a - b;
  if (ret < 0) ret += 1000000007;
  return ret;
}
int a[1007], n, i, dp[1007][300], pre[10][1007][1007], mn, mx, mid, vis[10], j,
    k, ans, mask, f;
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    a[i]--;
    vis[a[i]] = 1;
  }
  for (i = 0; i < 8; i++) ans += vis[i];
  for (i = 0; i < 8; i++) {
    for (j = 1; j <= n; j++) {
      pre[a[j]][j][1] = j;
      for (k = (i == a[j]) ? 2 : 1; k <= n; k++) {
        pre[i][j][k] = pre[i][j - 1][k - (i == a[j])];
      }
    }
  }
  mn = 1;
  mx = n;
  while (mn < mx) {
    ;
    mid = (mn + mx) / 2;
    if (mid == mn) mid++;
    ;
    for (i = 0; i <= n; i++)
      for (mask = 0; mask < 256; mask++) dp[i][mask] = -1;
    dp[0][0] = 0;
    for (i = 1; i <= n; i++) {
      for (mask = 0; mask < 256; mask++) {
        dp[i][mask] = dp[i - 1][mask];
        if (mask & (1 << a[i])) {
          if (pre[a[i]][i][mid]) {
            if (dp[pre[a[i]][i][mid] - 1][mask ^ (1 << a[i])] != -1)
              dp[i][mask] =
                  max(dp[pre[a[i]][i][mid] - 1][mask ^ (1 << a[i])] + mid,
                      dp[i][mask]);
            f = 1;
          }
          if (pre[a[i]][i][mid - 1]) {
            if (dp[pre[a[i]][i][mid - 1] - 1][mask ^ (1 << a[i])] != -1)
              dp[i][mask] = max(
                  dp[pre[a[i]][i][mid - 1] - 1][mask ^ (1 << a[i])] + mid - 1,
                  dp[i][mask]);
            f = 1;
          }
          if (dp[i][mask] != dp[i - 1][mask])
            ;
          if (dp[i][255] != -1)
            ;
        }
      }
    }
    if (mid == 2)
      ;
    if (dp[n][255] != -1)
      mn = mid;
    else
      mx = mid - 1;
    if (dp[n][255] != -1) ans = max(ans, dp[n][255]);
  }
  printf("%d\n", ans);
  return 0;
}
