#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e4 + 10;
int dp[MAXN], num[MAXN], a[110];
map<int, int> mp;
int len;
int main() {
  int n, t;
  while (scanf("%d%d", &n, &t) == 2) {
    int maxnum = 0, maxcnt = 0;
    mp.clear();
    memset(dp, 0, sizeof(dp));
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      mp[a[i]]++;
      if (mp[a[i]] > maxcnt) {
        maxcnt = mp[a[i]];
        maxnum = a[i];
      }
    }
    int cnt = 0;
    for (int i = 1; i <= min(n, t); i++) {
      for (int j = 1; j <= n; j++) num[++cnt] = a[j];
    }
    int val = 0;
    for (int i = 1; i <= cnt; i++) {
      int pos = upper_bound(dp, dp + val, num[i]) - dp;
      if (pos == val)
        dp[val++] = num[i];
      else
        dp[pos] = num[i];
    }
    if (t > n) val += (t - n) * maxcnt;
    printf("%d\n", val);
  }
  return 0;
}

### Prompt

Your task is to create 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 MAXN = 1e4 + 10;
int dp[MAXN], num[MAXN], a[110];
map<int, int> mp;
int len;
int main() {
  int n, t;
  while (scanf("%d%d", &n, &t) == 2) {
    int maxnum = 0, maxcnt = 0;
    mp.clear();
    memset(dp, 0, sizeof(dp));
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      mp[a[i]]++;
      if (mp[a[i]] > maxcnt) {
        maxcnt = mp[a[i]];
        maxnum = a[i];
      }
    }
    int cnt = 0;
    for (int i = 1; i <= min(n, t); i++) {
      for (int j = 1; j <= n; j++) num[++cnt] = a[j];
    }
    int val = 0;
    for (int i = 1; i <= cnt; i++) {
      int pos = upper_bound(dp, dp + val, num[i]) - dp;
      if (pos == val)
        dp[val++] = num[i];
      else
        dp[pos] = num[i];
    }
    if (t > n) val += (t - n) * maxcnt;
    printf("%d\n", val);
  }
  return 0;
}

```