#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const double eps = 1e-8;
const int inf = 2147483647;
const int maxn = 1e5 + 10;
int a[1110];
int n, k;
int ans[1110];
int cnt[3030];
int solve() {
  memset(cnt, 0, sizeof(cnt));
  for (int i = 0; i < n; i++) cnt[a[i]]++;
  int MAXcnt = 0;
  for (int i = 0; i < 3030; i++)
    if (cnt[i] > MAXcnt) MAXcnt = cnt[i];
  for (int i = 0; i < 1110; i++) ans[i] = inf;
  int tmp = 0;
  while (tmp < k) {
    tmp++;
    for (int i = 0; i < n; i++) *upper_bound(ans, ans + 1110, a[i]) = a[i];
    if (lower_bound(ans, ans + 1110, inf) - ans >= n) break;
  }
  int anslen = lower_bound(ans, ans + 1110, inf) - ans;
  return anslen + (k - tmp) * MAXcnt;
}
int main(void) {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int ans = solve();
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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. 

### Response

```cpp
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const double eps = 1e-8;
const int inf = 2147483647;
const int maxn = 1e5 + 10;
int a[1110];
int n, k;
int ans[1110];
int cnt[3030];
int solve() {
  memset(cnt, 0, sizeof(cnt));
  for (int i = 0; i < n; i++) cnt[a[i]]++;
  int MAXcnt = 0;
  for (int i = 0; i < 3030; i++)
    if (cnt[i] > MAXcnt) MAXcnt = cnt[i];
  for (int i = 0; i < 1110; i++) ans[i] = inf;
  int tmp = 0;
  while (tmp < k) {
    tmp++;
    for (int i = 0; i < n; i++) *upper_bound(ans, ans + 1110, a[i]) = a[i];
    if (lower_bound(ans, ans + 1110, inf) - ans >= n) break;
  }
  int anslen = lower_bound(ans, ans + 1110, inf) - ans;
  return anslen + (k - tmp) * MAXcnt;
}
int main(void) {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int ans = solve();
  printf("%d\n", ans);
  return 0;
}

```