#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
const int MAXN = 200;
const int MAXV = 400;
const int V = 310;
int N;
int T;
int A[MAXV];
int dp[MAXV][MAXV];
int res[MAXV][MAXV];
int tmp[MAXV][MAXV];
void mult(int a[MAXV][MAXV], int b[MAXV][MAXV]) {
  for (int i = 0; i < V; i++) {
    for (int j = i; j < V; j++) {
      tmp[i][j] = 0;
      for (int k = i; k <= j; k++) {
        tmp[i][j] = max(tmp[i][j], a[i][k] + b[k][j]);
      }
    }
  }
  for (int i = 0; i < V; i++) {
    for (int j = i; j < V; j++) {
      a[i][j] = tmp[i][j];
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> N;
  cin >> T;
  memset(dp, 0, sizeof(dp));
  memset(res, 0, sizeof(res));
  for (int t = 0; t < N; t++) {
    int v;
    cin >> v;
    for (int i = 0; i <= v; i++) {
      int cnt = 0;
      for (int j = i; j <= v; j++) {
        cnt = max(cnt, dp[i][j]);
      }
      dp[i][v] = max(dp[i][v], cnt + 1);
    }
  }
  while (T) {
    if (T % 2) {
      mult(res, dp);
    }
    mult(dp, dp);
    T /= 2;
  }
  int ans = 0;
  for (int i = 0; i < V; i++) {
    for (int j = i; j < V; j++) {
      ans = max(ans, res[i][j]);
    }
  }
  cout << ans << '\n';
  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;
const int INF = 1e9;
const int MAXN = 200;
const int MAXV = 400;
const int V = 310;
int N;
int T;
int A[MAXV];
int dp[MAXV][MAXV];
int res[MAXV][MAXV];
int tmp[MAXV][MAXV];
void mult(int a[MAXV][MAXV], int b[MAXV][MAXV]) {
  for (int i = 0; i < V; i++) {
    for (int j = i; j < V; j++) {
      tmp[i][j] = 0;
      for (int k = i; k <= j; k++) {
        tmp[i][j] = max(tmp[i][j], a[i][k] + b[k][j]);
      }
    }
  }
  for (int i = 0; i < V; i++) {
    for (int j = i; j < V; j++) {
      a[i][j] = tmp[i][j];
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> N;
  cin >> T;
  memset(dp, 0, sizeof(dp));
  memset(res, 0, sizeof(res));
  for (int t = 0; t < N; t++) {
    int v;
    cin >> v;
    for (int i = 0; i <= v; i++) {
      int cnt = 0;
      for (int j = i; j <= v; j++) {
        cnt = max(cnt, dp[i][j]);
      }
      dp[i][v] = max(dp[i][v], cnt + 1);
    }
  }
  while (T) {
    if (T % 2) {
      mult(res, dp);
    }
    mult(dp, dp);
    T /= 2;
  }
  int ans = 0;
  for (int i = 0; i < V; i++) {
    for (int j = i; j < V; j++) {
      ans = max(ans, res[i][j]);
    }
  }
  cout << ans << '\n';
  return 0;
}

```