#include <bits/stdc++.h>
using namespace std;
const int maxn = 105;
int n, T;
int a[maxn * maxn];
int dp[maxn * maxn];
int s[maxn * maxn];
int p, v;
int cnt[305];
int main() {
  while (cin >> n >> T) {
    memset(cnt, 0, sizeof(cnt));
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      cnt[a[i]]++;
    }
    for (int i = n + 1; i <= min(T, n) * n; i++) a[i] = a[i - n];
    int top = 0;
    int ans = 0, anspos;
    for (int i = 1; i <= min(T, n) * n; i++) {
      int L = 1, R = top + 1;
      while (L < R) {
        int M = L + (R - L) / 2;
        if (a[i] >= s[M])
          L = M + 1;
        else
          R = M;
      }
      dp[i] = L;
      ans = max(ans, dp[i]);
      if (L > top)
        s[++top] = a[i];
      else
        s[L] = min(s[L], a[i]);
    }
    int Max = 0;
    for (int i = 1; i <= 300; i++) Max = max(Max, cnt[i]);
    ans += Max * max(T - n, 0);
    cout << ans << endl;
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
const int maxn = 105;
int n, T;
int a[maxn * maxn];
int dp[maxn * maxn];
int s[maxn * maxn];
int p, v;
int cnt[305];
int main() {
  while (cin >> n >> T) {
    memset(cnt, 0, sizeof(cnt));
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      cnt[a[i]]++;
    }
    for (int i = n + 1; i <= min(T, n) * n; i++) a[i] = a[i - n];
    int top = 0;
    int ans = 0, anspos;
    for (int i = 1; i <= min(T, n) * n; i++) {
      int L = 1, R = top + 1;
      while (L < R) {
        int M = L + (R - L) / 2;
        if (a[i] >= s[M])
          L = M + 1;
        else
          R = M;
      }
      dp[i] = L;
      ans = max(ans, dp[i]);
      if (L > top)
        s[++top] = a[i];
      else
        s[L] = min(s[L], a[i]);
    }
    int Max = 0;
    for (int i = 1; i <= 300; i++) Max = max(Max, cnt[i]);
    ans += Max * max(T - n, 0);
    cout << ans << endl;
  }
  return 0;
}

```