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;
int cnt[305], a[105], dp[305];
int main() {
  int n, t;
  while (~scanf("%d%d", &n, &t)) {
    memset(cnt, 0, sizeof(cnt));
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      cnt[a[i]]++;
    }
    int mid = 0;
    for (int i = 0; i < n; i++) {
      if (cnt[a[i]] > mid) mid = cnt[a[i]];
    }
    for (int i = 0; i < min(t, 100); i++) {
      for (int j = 0; j < n; j++) {
        int maxx = 0;
        for (int k = 1; k <= a[j]; k++) {
          if (dp[k] > maxx) maxx = dp[k];
        }
        dp[a[j]] = maxx + 1;
      }
    }
    int ans = 0;
    for (int i = 0; i < n; i++) {
      if (dp[a[i]] > ans) ans = dp[a[i]];
    }
    if (t <= 100)
      printf("%d\n", ans);
    else
      printf("%d\n", ans + (t - 100) * mid);
  }
  return 0;
}
