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 = 301;
int n, T, a[MaxN], bit[MaxN], ans, res, f[MaxN];
void update(int x, int v) {
  for (; x < MaxN; x += x & -x) {
    bit[x] = max(bit[x], v);
  }
}
int query(int x) {
  int res = 0;
  for (; x > 0; x &= x - 1) res = max(res, bit[x]);
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> T;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int j = 1; j <= min(n, T); j++) {
    for (int i = 1; i <= n; i++) {
      f[i] = query(a[i]) + 1;
      update(a[i], f[i]);
    }
  }
  for (int i = 1; i <= n; i++) {
    ans = max(ans, f[i]);
  }
  if (n >= T) {
    cout << ans;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    f[i] = query(a[i]) + 1;
    update(a[i], f[i]);
    res = max(res, f[i]);
  }
  cout << ans + (T - n) * (res - ans);
  return 0;
}
