#include <bits/stdc++.h>
using namespace std;
int n, T;
int a[100005];
int b[100005];
int cnt[1000];
int dp[400];
int dp2[400];
int ans[400];
int main() {
  scanf("%d%d", &n, &T);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]), b[i] = a[i], cnt[a[i]]++;
  int id = n;
  int sz = n * min(n, T);
  for (int i = n; i < sz; i++) a[id] = a[id - n], id++;
  int sz2 = n * min(n, max(0, T - n));
  id = n;
  for (int i = n; i < sz2; i++) b[id] = b[id - n], id++;
  int last = max(0, T - 2 * n);
  for (int i = 0; i < sz; i++) {
    int ma = 0;
    for (int j = 0; j <= a[i]; j++) ma = max(ma, dp[j]);
    dp[a[i]] = max(dp[a[i]], ma + 1);
  }
  for (int i = sz2 - 1; i >= 0; i--) {
    int ma = 0;
    for (int j = b[i]; j <= 300; j++) ma = max(ma, dp2[j]);
    dp2[b[i]] = max(dp2[b[i]], ma + 1);
  }
  int ma = 0;
  for (int i = 0; i <= 300; i++) {
    ma = max(ma, dp[i]);
    ans[i] = ma + cnt[i] * last;
  }
  ma = 0;
  int res = 0;
  for (int i = 300; i >= 0; i--) {
    ma = max(ma, dp2[i]);
    res = max(res, ans[i] + ma);
  }
  cout << res << endl;
  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>
using namespace std;
int n, T;
int a[100005];
int b[100005];
int cnt[1000];
int dp[400];
int dp2[400];
int ans[400];
int main() {
  scanf("%d%d", &n, &T);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]), b[i] = a[i], cnt[a[i]]++;
  int id = n;
  int sz = n * min(n, T);
  for (int i = n; i < sz; i++) a[id] = a[id - n], id++;
  int sz2 = n * min(n, max(0, T - n));
  id = n;
  for (int i = n; i < sz2; i++) b[id] = b[id - n], id++;
  int last = max(0, T - 2 * n);
  for (int i = 0; i < sz; i++) {
    int ma = 0;
    for (int j = 0; j <= a[i]; j++) ma = max(ma, dp[j]);
    dp[a[i]] = max(dp[a[i]], ma + 1);
  }
  for (int i = sz2 - 1; i >= 0; i--) {
    int ma = 0;
    for (int j = b[i]; j <= 300; j++) ma = max(ma, dp2[j]);
    dp2[b[i]] = max(dp2[b[i]], ma + 1);
  }
  int ma = 0;
  for (int i = 0; i <= 300; i++) {
    ma = max(ma, dp[i]);
    ans[i] = ma + cnt[i] * last;
  }
  ma = 0;
  int res = 0;
  for (int i = 300; i >= 0; i--) {
    ma = max(ma, dp2[i]);
    res = max(res, ans[i] + ma);
  }
  cout << res << endl;
  return 0;
}

```