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;
const double eps = 1e-7;
int ary[105];
int BIT[310];
void update(int a[], int idx, int v) {
  for (int i = idx + 5; i < 310; i += i & -i) a[i] = max(a[i], v);
}
int query(int a[], int idx) {
  int ans = 0;
  for (int i = idx + 5; i > 0; i -= i & -i) ans = max(a[i], ans);
  return ans;
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; ++i) scanf("%d", &ary[i]);
  vector<int> nums(ary, ary + n);
  (sort((nums).begin(), (nums).end()),
   ((nums).erase(unique((nums).begin(), (nums).end()), (nums).end())));
  memset(BIT, 0, sizeof(BIT));
  for (int k = 0; k < 100 and k < m; ++k) {
    for (int i = 0; i < n; ++i) {
      int v = query(BIT, ary[i]);
      update(BIT, ary[i], v + 1);
    }
  }
  int d = (m - 100 - 1) - (100 - 1);
  d = max(0, d);
  for (int i = ((int)(nums).size()) - 1; i >= 0; --i) {
    int a = nums[i];
    int cnt = 0;
    for (int j = 0; j < n; ++j)
      if (ary[j] == a) cnt++;
    int v = query(BIT, a);
    update(BIT, a, v + cnt * d);
  }
  for (int k = max(m - 100, 100); k < m; ++k) {
    for (int i = 0; i < n; ++i) {
      int v = query(BIT, ary[i]);
      update(BIT, ary[i], v + 1);
    }
  }
  int ans = query(BIT, 300);
  printf("%d\n", ans);
}
