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 = 1000100;
const int INF = (1 << 29);
int n, T;
int a[maxn];
int b[maxn], len;
int cnt[maxn];
int lis(int *a, int n) {
  len = 0;
  b[++len] = a[1];
  for (int i = 2; i <= n; i++) {
    if (a[i] >= b[len])
      b[++len] = a[i];
    else {
      int p = upper_bound(b + 1, b + len + 1, a[i]) - b;
      b[p] = a[i];
    }
  }
  return len;
}
int main() {
  cin >> n >> T;
  memset(cnt, 0, sizeof(cnt));
  int max_cnt = 0;
  for (int i = 1; i <= n; i++)
    scanf("%I64d", &a[i]), cnt[a[i]]++, max_cnt = max(max_cnt, cnt[a[i]]);
  for (int i = 1; i <= n - 1; i++) {
    for (int j = 1; j <= n; j++) {
      a[i * n + j] = a[j];
    }
  }
  long long ans = 0;
  if (T <= n)
    ans = lis(a, n * T);
  else {
    ans = lis(a, n * n) + max_cnt * (T - n);
  }
  printf("%I64d\n", ans);
  return 0;
}
