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. 
#include <bits/stdc++.h>
using namespace std;
const int N = 220;
long long temp[N], A[N * N], calc[2][305], dp[N * N], freq[305];
int main() {
  long long sz, cntr = 1, n, T, ans = 0;
  cin >> n >> T;
  for (int i = 1; i <= n; i++) {
    cin >> temp[i];
    freq[temp[i]]++;
  }
  if (T <= 2 * n) {
    for (int i = 1; i <= T; i++)
      for (int j = 1; j <= n; j++) A[cntr++] = temp[j];
    for (int i = cntr - 1; i >= 1; i--) {
      dp[i] = calc[0][A[i]] + 1;
      ans = max(ans, dp[i]);
      for (int j = A[i]; j >= 1; j--) calc[0][j] = max(calc[0][j], dp[i]);
    }
    cout << ans << endl;
  } else {
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) A[cntr++] = temp[j];
    memset(dp, 0, sizeof(dp));
    for (int i = 1; i < cntr; i++) {
      dp[i] = calc[0][A[i]] + 1;
      for (int j = A[i]; j <= 300; j++) calc[0][j] = max(calc[0][j], dp[i]);
    }
    memset(dp, 0, sizeof(dp));
    for (int i = cntr - 1; i >= 1; i--) {
      dp[i] = calc[1][A[i]] + 1;
      for (int j = A[i]; j >= 1; j--) calc[1][j] = max(calc[1][j], dp[i]);
    }
    for (int i = 1; i <= 300; i++)
      for (int j = i; j <= 300; j++)
        ans = max(ans, calc[0][i] + (freq[i] * (T - (2 * n))) + calc[1][j]);
    cout << ans << endl;
  }
}
