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, T, ans, a[100007], dp[100007];
vector<int> v[107];
int main() {
  scanf("%d %d", &n, &t);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  T = (107 < t ? 107 : t);
  t -= T;
  for (int i = 0; i < n; ++i) {
    for (int j = 1; j < T; ++j) a[i + (j * n)] = a[i];
  }
  dp[0] = 1;
  v[0].push_back(1);
  for (int i = 1; i < (T * n); ++i) {
    dp[i] = 1;
    for (int j = 0; j < i; ++j) {
      if (a[j] <= a[i]) dp[i] = (dp[i] > dp[j] + 1 ? dp[i] : dp[j] + 1);
    }
    v[i % n].push_back(dp[i]);
  }
  if (T == 1) {
    ans = -1;
    for (int i = 0; i < n; ++i) ans = (ans > dp[i] ? ans : dp[i]);
    printf("%d", ans);
    return 0;
  } else {
    ans = -1;
    for (int i = 0; i < n; ++i) {
      ans = (ans > v[i][v[i].size() - 1] +
                         (t * (v[i][v[i].size() - 1] - v[i][v[i].size() - 2]))
                 ? ans
                 : v[i][v[i].size() - 1] +
                       (t * (v[i][v[i].size() - 1] - v[i][v[i].size() - 2])));
    }
    printf("%d", ans);
  }
  return 0;
}
