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 = 105;
const int M = 1000000007;
int n, t;
int a[N];
int b[N * 100];
inline int get(int i) { return a[i % n]; }
int main() {
  while (~scanf("%d%d", &n, &t)) {
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
    }
    int tt = min(100, t), len = 1;
    b[0] = a[0];
    for (int i = 1; i < n * tt; i++) {
      int j =
          get(i) >= b[len - 1] ? len++ : upper_bound(b, b + len, get(i)) - b;
      b[j] = get(i);
    }
    if (t <= tt) {
      printf("%d\n", len);
      continue;
    }
    int len2 = 1;
    b[0] = a[0];
    for (int i = 1; i < n * (tt + 1); i++) {
      int j =
          get(i) >= b[len2 - 1] ? len2++ : upper_bound(b, b + len2, get(i)) - b;
      b[j] = get(i);
    }
    printf("%d\n", len + (len2 - len) * (t - tt));
  }
}
