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 INF = 1e9 + 100;
int a[105];
int n, t;
vector<vector<int>> mult(vector<vector<int>> x, vector<vector<int>> y) {
  vector<vector<int>> ret(n + 1, vector<int>(n + 1, -INF));
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      for (int k = 1; k <= n; ++k) {
        ret[i][j] = max(ret[i][j], x[i][k] + y[k][j]);
      }
    }
  }
  return ret;
}
vector<vector<int>> pow(vector<vector<int>> m, int b) {
  vector<vector<int>> x = m;
  vector<vector<int>> ret(n + 1, vector<int>(n + 1));
  while (b) {
    if (b & 1) {
      ret = mult(ret, x);
    }
    x = mult(x, x);
    b >>= 1;
  }
  return ret;
}
int main() {
  scanf("%d%d", &n, &t);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  vector<vector<int>> dp(n + 1, vector<int>(n + 1, -INF));
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (a[i] > a[j]) continue;
      dp[i][j] = 1;
      for (int k = 0; k < j; ++k) {
        if (a[k] <= a[j]) dp[i][j] = max(dp[i][j], dp[i][k] + 1);
      }
    }
  }
  vector<vector<int>> res = pow(dp, t);
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      ans = max(ans, res[i][j]);
    }
  }
  printf("%lld\n", ans);
}
