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 OK = 1;
const int INF = 0x3f3f3f3f;
const int maxn = 400300;
int arr[maxn], f[maxn];
int mx = 300;
void add(int i, int v) {
  for (; i <= mx; i += i & (-i))
    if (f[i] < v) f[i] = v;
}
int query(int i) {
  int ret = 0;
  for (; i; i -= i & (-i))
    if (ret < f[i]) ret = f[i];
  return ret;
}
int num[330];
int main() {
  int n, T;
  scanf("%d%d", &n, &T);
  for (int i = 1; i <= n; ++i) scanf("%d", arr + i);
  for (int i = 1; i <= n; ++i) {
    num[arr[i]]++;
  }
  int cnt = n + 1;
  if (T <= 4000) {
    for (int i = 2; i <= T; ++i) {
      for (int j = 1; j <= n; ++j) arr[cnt++] = arr[j];
    }
  } else {
    for (int i = 2; i <= 4000; ++i)
      for (int j = 1; j <= n; ++j) arr[cnt++] = arr[j];
  }
  int mx = 0;
  for (int i = 1; i < cnt; ++i) {
    int t = query(arr[i]);
    t++;
    add(arr[i], t);
    if (t > mx) mx = t;
  }
  if (T <= 4000) {
    cout << mx << endl;
  } else {
    int mm = 0;
    for (int i = 1; i <= 300; ++i)
      if (num[i] > mm) mm = num[i];
    cout << mx + (T - 4000) * mm << endl;
  }
  return 0;
}
