#include <bits/stdc++.h>
const int N = 100 + 10;
int num[N * N];
int dp[N * N];
int hash[500];
inline int Max(int a, int b) { return a > b ? a : b; }
inline int Min(int a, int b) { return a < b ? a : b; }
int lis(int num[], int n) {
  int ans = -1;
  memset(dp, 0, sizeof(dp));
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < i; j++) {
      if (num[j] <= num[i]) dp[i] = Max(dp[j] + 1, dp[i]);
      ans = Max(ans, dp[i]);
    }
  }
  return ans;
}
int main(void) {
  int n, T;
  memset(hash, 0, sizeof(hash));
  scanf("%d%d", &n, &T);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &num[i]);
    hash[num[i]] += 1;
  }
  int ans = 0;
  for (int i = 0, times = Min(T, n); i < times; i++) {
    for (int j = 1; j <= n; j++) {
      num[i * n + j] = num[j];
    }
  }
  ans = lis(num, Min(n, T) * n);
  int max = 0;
  for (int i = 1; i <= 300; i++) {
    max = Max(max, hash[i]);
  }
  ans += max * (T - Min(n, T));
  printf("%d\n", ans);
}

### Prompt

Develop a solution in Cpp to the problem described below:

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>
const int N = 100 + 10;
int num[N * N];
int dp[N * N];
int hash[500];
inline int Max(int a, int b) { return a > b ? a : b; }
inline int Min(int a, int b) { return a < b ? a : b; }
int lis(int num[], int n) {
  int ans = -1;
  memset(dp, 0, sizeof(dp));
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < i; j++) {
      if (num[j] <= num[i]) dp[i] = Max(dp[j] + 1, dp[i]);
      ans = Max(ans, dp[i]);
    }
  }
  return ans;
}
int main(void) {
  int n, T;
  memset(hash, 0, sizeof(hash));
  scanf("%d%d", &n, &T);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &num[i]);
    hash[num[i]] += 1;
  }
  int ans = 0;
  for (int i = 0, times = Min(T, n); i < times; i++) {
    for (int j = 1; j <= n; j++) {
      num[i * n + j] = num[j];
    }
  }
  ans = lis(num, Min(n, T) * n);
  int max = 0;
  for (int i = 1; i <= 300; i++) {
    max = Max(max, hash[i]);
  }
  ans += max * (T - Min(n, T));
  printf("%d\n", ans);
}

```