#include <bits/stdc++.h>
using namespace std;
const int B = 40, inf = 2e9 + 7;
int F[B][B], M[B][B], z[B][B];
int dp[B][B][B];
int a[B], b[B];
int bs;
void multi(int x[B][B], int y[B][B]) {
  int i, j, k;
  for (i = 0; i < bs; i++)
    for (j = 0; j < bs; j++) z[i][j] = inf;
  for (i = 0; i < bs; i++)
    for (j = 0; j < bs; j++)
      for (k = 0; k < bs; k++)
        if (1ll * x[i][k] + y[k][j] < inf)
          z[i][j] = min(z[i][j], x[i][k] + y[k][j]);
  for (i = 0; i < bs; i++)
    for (j = 0; j < bs; j++) x[i][j] = z[i][j];
}
void pow(int n) {
  if (n == 1) return;
  pow(n / 2);
  multi(F, F);
  if (n & 1) multi(F, M);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m, i, j, k, d;
  cin >> n >> m;
  bs = 2 * n;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++) cin >> b[i];
  for (i = 0; i < B; i++)
    for (j = 0; j < B; j++)
      for (k = 0; k < B; k++) dp[i][j][k] = inf;
  for (d = 0; d < 2 * n; d++) {
    dp[d][0][d] = 0;
    for (i = 1; i <= n; i++) {
      dp[d][i][0] = dp[d][i - 1][1] + b[i];
      dp[d][i][2 * n - 1] = dp[d][i - 1][2 * n - 2] + a[i];
      for (j = 1; j < 2 * n - 1; j++)
        dp[d][i][j] =
            min(dp[d][i - 1][j - 1] + a[i], dp[d][i - 1][j + 1] + b[i]);
    }
  }
  for (i = 0; i < bs; i++)
    for (j = 0; j < bs; j++) M[i][j] = F[i][j] = dp[j][n][i];
  pow(m);
  cout << F[0][0];
}

### Prompt

Create a solution in CPP for 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 int B = 40, inf = 2e9 + 7;
int F[B][B], M[B][B], z[B][B];
int dp[B][B][B];
int a[B], b[B];
int bs;
void multi(int x[B][B], int y[B][B]) {
  int i, j, k;
  for (i = 0; i < bs; i++)
    for (j = 0; j < bs; j++) z[i][j] = inf;
  for (i = 0; i < bs; i++)
    for (j = 0; j < bs; j++)
      for (k = 0; k < bs; k++)
        if (1ll * x[i][k] + y[k][j] < inf)
          z[i][j] = min(z[i][j], x[i][k] + y[k][j]);
  for (i = 0; i < bs; i++)
    for (j = 0; j < bs; j++) x[i][j] = z[i][j];
}
void pow(int n) {
  if (n == 1) return;
  pow(n / 2);
  multi(F, F);
  if (n & 1) multi(F, M);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m, i, j, k, d;
  cin >> n >> m;
  bs = 2 * n;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++) cin >> b[i];
  for (i = 0; i < B; i++)
    for (j = 0; j < B; j++)
      for (k = 0; k < B; k++) dp[i][j][k] = inf;
  for (d = 0; d < 2 * n; d++) {
    dp[d][0][d] = 0;
    for (i = 1; i <= n; i++) {
      dp[d][i][0] = dp[d][i - 1][1] + b[i];
      dp[d][i][2 * n - 1] = dp[d][i - 1][2 * n - 2] + a[i];
      for (j = 1; j < 2 * n - 1; j++)
        dp[d][i][j] =
            min(dp[d][i - 1][j - 1] + a[i], dp[d][i - 1][j + 1] + b[i]);
    }
  }
  for (i = 0; i < bs; i++)
    for (j = 0; j < bs; j++) M[i][j] = F[i][j] = dp[j][n][i];
  pow(m);
  cout << F[0][0];
}

```