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 int maxn = 5010;
int f[maxn];
int a[maxn];
int num[maxn];
int n, t;
int ans = 0;
int lim;
int maxnum = 0;
int main() {
  scanf("%d%d", &n, &t);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    num[a[i]]++;
    maxnum = max(maxnum, num[a[i]]);
  }
  lim = min(n / 2, t) * n;
  if (n == 1) {
    printf("%d", t);
    return 0;
  }
  for (int i = n + 1; i <= lim; i++) {
    a[i] = a[i - n];
  }
  for (int i = 1; i <= lim; i++) f[i] = 1;
  for (int i = 1; i <= lim; i++) {
    for (int j = 1; j < i; j++) {
      if (a[j] <= a[i]) {
        f[i] = max(f[i], f[j] + 1);
      }
    }
  }
  for (int i = 1; i <= lim; i++) ans = max(ans, f[i]);
  if (t <= (n / 2))
    printf("%d", ans);
  else
    printf("%d", ans + (t - (n / 2)) * maxnum);
  return 0;
}
