You are given an array of positive integers a1, a2, ..., an × T of length n × T. We know that for any i > n it is true that ai = ai - n. Find the length of the longest non-decreasing sequence of the given array.

Input

The first line contains two space-separated integers: n, T (1 ≤ n ≤ 100, 1 ≤ T ≤ 107). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 300).

Output

Print a single number — the length of a sought sequence.

Examples

Input

4 3
3 1 4 2


Output

5

Note

The array given in the sample looks like that: 3, 1, 4, 2, 3, 1, 4, 2, 3, 1, 4, 2. The elements in bold form the largest non-decreasing subsequence. 
#include <bits/stdc++.h>
using namespace std;
int n, T, f[222][111], g[111][111], a[111], cnt[333];
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> T;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  int m = min(n, T);
  for (int i = 1; i <= m * 2; i++)
    for (int j = 1; j <= n; j++) {
      f[i][j] = 1;
      for (int k = 1; k <= n; k++)
        if (a[k] <= a[j]) {
          if (k < j) f[i][j] = max(f[i][j], f[i][k] + 1);
          f[i][j] = max(f[i][j], f[i - 1][k] + 1);
        }
    }
  for (int i = 1; i <= m; i++)
    for (int j = n; j; j--) {
      g[i][j] = 1;
      for (int k = 1; k <= n; k++)
        if (a[j] <= a[k]) {
          if (k > j) g[i][j] = max(g[i][j], g[i][k] + 1);
          g[i][j] = max(g[i][j], g[i - 1][k] + 1);
        }
    }
  int ans = 0;
  if (T <= m * 2) {
    for (int i = 1; i <= n; i++) ans = max(ans, f[T][i]);
  } else {
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++)
        for (int k = 1; k <= n; k++)
          if (a[i] <= a[j] && a[j] <= a[k])
            ans = max(ans, f[m][i] + g[m][k] + (T - m * 2) * cnt[a[j]]);
  }
  cout << ans << endl;
}
