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 = 105;
const int maxm = 305;
int c[maxm];
int a[maxn], b[maxn * maxn * 2];
int dp[maxm];
int suf[maxn * maxn], pre[maxn * maxn];
int main() {
  ios::sync_with_stdio(false);
  int n, t;
  int mx;
  int ans = 0;
  scanf("%d %d", &n, &t);
  for (int i = 1; i < n + 1; ++i) {
    scanf("%d", &a[i]);
    ++c[a[i]];
  }
  if (t <= n * 2) {
    for (int i = 1; i < n + 1; ++i) {
      int k = i;
      for (int j = 1; j < t + 1; ++j) {
        b[k] = a[i];
        k += n;
      }
    }
    int n_ = n * t;
    memset(dp, 0, sizeof(dp));
    for (int i = 1; i < n_ + 1; ++i) {
      mx = 0;
      for (int j = 1; j < b[i] + 1; ++j) mx = max(mx, dp[j]);
      pre[i] = ++mx;
      dp[b[i]] = mx;
    }
    for (int i = 1; i < n_ + 1; ++i) ans = max(ans, pre[i]);
    printf("%d\n", ans);
    return 0;
  }
  for (int i = 1; i < n + 1; ++i) {
    int k = i;
    for (int j = 1; j < n + 1; ++j) {
      b[k] = a[i];
      k += n;
    }
  }
  int n_ = n * n;
  memset(dp, 0, sizeof(dp));
  for (int i = 1; i < n_ + 1; ++i) {
    mx = 0;
    for (int j = 1; j < b[i] + 1; ++j) mx = max(mx, dp[j]);
    pre[i] = ++mx;
    dp[b[i]] = mx;
  }
  memset(dp, 0, sizeof(dp));
  for (int i = n_ + 1 - 1; i >= 1; --i) {
    mx = 0;
    for (int j = b[i]; j < maxm; ++j) mx = max(mx, dp[j]);
    suf[i] = ++mx;
    dp[b[i]] = mx;
  }
  int tmp, n2 = n + n;
  for (int i = 1; i < n + 1; ++i) {
    for (int j = 1; j < n + 1; ++j) {
      if (a[j] < a[i]) continue;
      tmp = pre[i + n_ - n] + suf[j] + c[a[i]] * (t - n2);
      ans = max(ans, tmp);
    }
  }
  printf("%d\n", ans);
  return 0;
}
