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 long long INF = 1000000000;
const double esp = 1e-13;
int n, t, a[100 + 10];
struct BIG {
  int d[150][150];
  BIG(int v = 0) { memset(d, 0, sizeof(d)); }
  BIG operator*(const BIG &op) const {
    BIG res;
    for (long long i = 1; i <= n; i++)
      for (long long j = 1; j <= n; j++) {
        res.d[i][j] = -INF;
        for (long long k = 1; k <= n; k++)
          res.d[i][j] = max(res.d[i][j], d[i][k] + op.d[k][j]);
      }
    return res;
  }
};
BIG f, ans;
int answer;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  srand(time(0));
  cin >> n >> t;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= n; j++)
      if (a[i] < a[j])
        f.d[j][i] = -INF;
      else {
        f.d[j][i] = 1;
        for (long long k = 1; k <= i - 1; k++)
          if (a[i] >= a[k]) f.d[j][i] = max(f.d[j][i], f.d[j][k] + 1);
      }
  ans = f;
  t--;
  while (t) {
    if (t % 2) ans = ans * f;
    f = f * f;
    t /= 2;
  }
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= n; j++) answer = max(answer, ans.d[i][j]);
  cout << answer;
}
