#include <bits/stdc++.h>
using namespace std;
long n, t, a[1000000 + 1], b[1000000 + 1], kq, f[1000000 + 1], bit[400];
void doc() {
  cin >> n >> t;
  for (int i = 1; i <= n; i++) cin >> a[i];
}
long get(long k) {
  long ans = 0;
  while (k > 0) {
    ans = max(ans, bit[k]);
    k = k - (k & (-k));
  }
  return ans;
}
void update(long k, long x) {
  while (k <= 300) {
    bit[k] = max(bit[k], x);
    k += k & (-k);
  }
}
long tinh(long k) {
  long x, m = k * n;
  for (int j = 1; j <= k; j++)
    for (int i = 1; i <= n; i++) b[(j - 1) * (n) + i] = a[i];
  for (int i = 1; i <= 300; i++) bit[i] = 0;
  for (int i = 1; i <= m; i++) {
    x = get(b[i]) + 1;
    update(b[i], x);
    f[i] = max(x, f[i - 1]);
  }
  return f[m];
}
void xl() {
  if (n * t <= 1000000) {
    cout << tinh(t);
  } else {
    kq = (tinh(102) - tinh(101)) * (t - 101) + tinh(101);
    cout << kq;
  }
}
int main() {
  doc();
  xl();
}

### Prompt

Generate 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;
long n, t, a[1000000 + 1], b[1000000 + 1], kq, f[1000000 + 1], bit[400];
void doc() {
  cin >> n >> t;
  for (int i = 1; i <= n; i++) cin >> a[i];
}
long get(long k) {
  long ans = 0;
  while (k > 0) {
    ans = max(ans, bit[k]);
    k = k - (k & (-k));
  }
  return ans;
}
void update(long k, long x) {
  while (k <= 300) {
    bit[k] = max(bit[k], x);
    k += k & (-k);
  }
}
long tinh(long k) {
  long x, m = k * n;
  for (int j = 1; j <= k; j++)
    for (int i = 1; i <= n; i++) b[(j - 1) * (n) + i] = a[i];
  for (int i = 1; i <= 300; i++) bit[i] = 0;
  for (int i = 1; i <= m; i++) {
    x = get(b[i]) + 1;
    update(b[i], x);
    f[i] = max(x, f[i - 1]);
  }
  return f[m];
}
void xl() {
  if (n * t <= 1000000) {
    cout << tinh(t);
  } else {
    kq = (tinh(102) - tinh(101)) * (t - 101) + tinh(101);
    cout << kq;
  }
}
int main() {
  doc();
  xl();
}

```