#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void chkmin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline void chkmax(T &a, T b) {
  if (a < b) a = b;
}
int N, a[30], b[30], f[30][30], g[30][30];
void multi(int f[30][30], int g[30][30]) {
  int tmp[30][30], i, j, k;
  for (i = 0; i <= N; i++)
    for (j = 0; j <= N; j++) {
      tmp[i][j] = 2100000000;
      for (k = 0; k <= N; k++)
        if (f[i][k] < 2100000000 && g[k][j] < 2100000000)
          chkmin(tmp[i][j], f[i][k] + g[k][j]);
    }
  for (i = 0; i <= N; i++)
    for (j = 0; j <= N; j++) f[i][j] = tmp[i][j];
}
int main() {
  int M, i, j, dp, md, sum, u, v;
  scanf("%d%d", &N, &M);
  for (i = 0; i < N; i++) scanf("%d", &a[i]);
  for (i = 0; i < N; i++) scanf("%d", &b[i]);
  for (i = 0; i <= N; i++) {
    for (j = 0; j <= N; j++) f[i][j] = g[i][j] = 2100000000;
    f[i][i] = 0;
  }
  for (i = 0; i < (1 << N); i++) {
    dp = 0;
    md = 0;
    sum = 0;
    for (j = 0; j < N; j++) {
      if (i & (1 << j)) {
        sum += a[j];
        dp++;
      } else {
        sum += b[j];
        dp--;
      }
      chkmin(md, dp);
    }
    for (j = -md; j <= N; j++)
      if (j + dp >= 0 && j + dp <= N) {
        u = j;
        v = j + dp;
        chkmin(g[u][v], sum);
      }
  }
  for (i = 30; i >= 0; i--) {
    multi(f, f);
    if (M & (1 << i)) {
      multi(f, g);
    }
  }
  printf("%d\n", f[0][0]);
  return 0;
}

### Prompt

Your task is to create 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;
template <class T>
inline void chkmin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline void chkmax(T &a, T b) {
  if (a < b) a = b;
}
int N, a[30], b[30], f[30][30], g[30][30];
void multi(int f[30][30], int g[30][30]) {
  int tmp[30][30], i, j, k;
  for (i = 0; i <= N; i++)
    for (j = 0; j <= N; j++) {
      tmp[i][j] = 2100000000;
      for (k = 0; k <= N; k++)
        if (f[i][k] < 2100000000 && g[k][j] < 2100000000)
          chkmin(tmp[i][j], f[i][k] + g[k][j]);
    }
  for (i = 0; i <= N; i++)
    for (j = 0; j <= N; j++) f[i][j] = tmp[i][j];
}
int main() {
  int M, i, j, dp, md, sum, u, v;
  scanf("%d%d", &N, &M);
  for (i = 0; i < N; i++) scanf("%d", &a[i]);
  for (i = 0; i < N; i++) scanf("%d", &b[i]);
  for (i = 0; i <= N; i++) {
    for (j = 0; j <= N; j++) f[i][j] = g[i][j] = 2100000000;
    f[i][i] = 0;
  }
  for (i = 0; i < (1 << N); i++) {
    dp = 0;
    md = 0;
    sum = 0;
    for (j = 0; j < N; j++) {
      if (i & (1 << j)) {
        sum += a[j];
        dp++;
      } else {
        sum += b[j];
        dp--;
      }
      chkmin(md, dp);
    }
    for (j = -md; j <= N; j++)
      if (j + dp >= 0 && j + dp <= N) {
        u = j;
        v = j + dp;
        chkmin(g[u][v], sum);
      }
  }
  for (i = 30; i >= 0; i--) {
    multi(f, f);
    if (M & (1 << i)) {
      multi(f, g);
    }
  }
  printf("%d\n", f[0][0]);
  return 0;
}

```