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;
long long gcd(long long a, long long b) {
  long long r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const int maxn = 110;
int n, t;
int a[maxn];
int cnt[3 * maxn];
int f[2 * maxn * maxn];
int d[2 * maxn * maxn];
int g[3 * maxn];
int calc(int h) {
  if (g[h] != -1) return g[h];
  int res = 0;
  memset(d, 0x3f, sizeof(d));
  for (int i = (0); i < (n * n); i++)
    if (a[i % n] >= h) {
      int k = upper_bound(d, d + n * n, a[i % n]) - d;
      d[k] = a[i % n];
      res = max(res, k + 1);
    }
  return g[h] = res;
}
void solve() {
  memset(cnt, 0, sizeof(cnt));
  memset(f, 0, sizeof(f));
  memset(g, -1, sizeof(g));
  scanf("%d%d", &n, &t);
  for (int i = (0); i < (n); i++) {
    scanf("%d", a + i);
    cnt[a[i]]++;
  }
  if (t <= 2 * n) {
    int ans = 0;
    memset(d, 0x3f, sizeof(d));
    for (int i = (0); i < (t * n); i++) {
      int k = upper_bound(d, d + n * n, a[i % n]) - d;
      d[k] = a[i % n];
      ans = max(ans, k + 1);
    }
    printf("%d", ans);
    return;
  }
  memset(d, 0x3f, sizeof(d));
  for (int i = (0); i < (n * n); i++) {
    int k = upper_bound(d, d + n * n, a[i % n]) - d;
    d[k] = a[i % n];
    f[i] = k + 1;
  }
  int ans = 0;
  for (int i = (n * n - n); i < (n * n); i++) {
    ans = max(ans, f[i] + (t - 2 * n) * cnt[a[i % n]] + calc(a[i % n]));
  }
  printf("%d", ans);
}
int main() {
  solve();
  return 0;
}
