#include <bits/stdc++.h>
using namespace std;
const long long maxN = 1e7 + 5, maxM = 45, mod = 1e9 + 7;
long long N, M;
long long a[maxM], b[maxM], c[maxM];
long long cost;
long long s[maxM];
bool b1[maxM], b0[maxM];
bool cmp(long long i, long long j) { return c[i] < c[j]; }
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> M >> N;
  long long tot = M * N;
  for (long long i = (long long)0; i <= (long long)M - 1; i++) {
    cin >> a[i];
    a[i + M] = a[i];
  }
  for (long long i = (long long)0; i <= (long long)M - 1; i++) {
    cin >> b[i];
    b[i + M] = b[i];
  }
  M *= 2;
  N /= 2;
  for (long long i = (long long)0; i <= (long long)M - 1; i++) {
    cost += b[i] * N;
    c[i] = a[i] - b[i];
    s[i] = i;
  }
  sort(s, s + M, cmp);
  for (long long rnk = (long long)0; rnk <= (long long)M - 1; rnk++) {
    long long i = s[rnk];
    for (long long j = i + (i & 1); j < M; j += 2) {
      if (!b0[j]) {
        b0[j] = 1;
        cost += c[i];
        break;
      }
    }
    for (long long j = 0; j < M; j += 2) {
      if (!b1[j]) {
        b1[j] = 1;
        cost += c[i] * (N - 1);
        break;
      }
    }
  }
  cout << cost << '\n';
}

### Prompt

Please formulate a cpp solution to the following problem:

Jeff loves regular bracket sequences.

Today Jeff is going to take a piece of paper and write out the regular bracket sequence, consisting of nm brackets. Let's number all brackets of this sequence from 0 to nm - 1 from left to right. Jeff knows that he is going to spend ai mod n liters of ink on the i-th bracket of the sequence if he paints it opened and bi mod n liters if he paints it closed.

You've got sequences a, b and numbers n, m. What minimum amount of ink will Jeff need to paint a regular bracket sequence of length nm?

Operation x mod y means taking the remainder after dividing number x by number y.

Input

The first line contains two integers n and m (1 ≤ n ≤ 20; 1 ≤ m ≤ 107; m is even). The next line contains n integers: a0, a1, ..., an - 1 (1 ≤ ai ≤ 10). The next line contains n integers: b0, b1, ..., bn - 1 (1 ≤ bi ≤ 10). The numbers are separated by spaces.

Output

In a single line print the answer to the problem — the minimum required amount of ink in liters.

Examples

Input

2 6
1 2
2 1


Output

12


Input

1 10000000
2
3


Output

25000000

Note

In the first test the optimal sequence is: ()()()()()(), the required number of ink liters is 12.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long maxN = 1e7 + 5, maxM = 45, mod = 1e9 + 7;
long long N, M;
long long a[maxM], b[maxM], c[maxM];
long long cost;
long long s[maxM];
bool b1[maxM], b0[maxM];
bool cmp(long long i, long long j) { return c[i] < c[j]; }
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> M >> N;
  long long tot = M * N;
  for (long long i = (long long)0; i <= (long long)M - 1; i++) {
    cin >> a[i];
    a[i + M] = a[i];
  }
  for (long long i = (long long)0; i <= (long long)M - 1; i++) {
    cin >> b[i];
    b[i + M] = b[i];
  }
  M *= 2;
  N /= 2;
  for (long long i = (long long)0; i <= (long long)M - 1; i++) {
    cost += b[i] * N;
    c[i] = a[i] - b[i];
    s[i] = i;
  }
  sort(s, s + M, cmp);
  for (long long rnk = (long long)0; rnk <= (long long)M - 1; rnk++) {
    long long i = s[rnk];
    for (long long j = i + (i & 1); j < M; j += 2) {
      if (!b0[j]) {
        b0[j] = 1;
        cost += c[i];
        break;
      }
    }
    for (long long j = 0; j < M; j += 2) {
      if (!b1[j]) {
        b1[j] = 1;
        cost += c[i] * (N - 1);
        break;
      }
    }
  }
  cout << cost << '\n';
}

```