Problem: Polycarp is a regular customer at the restaurant "Ber Patio". He likes having lunches there.

"Ber Patio" has special discount program for regular customers. A customer can collect bonuses and partially cover expenses in the restaurant.

Let's assume a customer currently has b bonuses and she has to pay r burles for a lunch. In this case the customer can use bonuses (1 bonus = 1 burle) to reduce the payment. She can cover at most half of the payment using bonuses. However, 1 bonus will be added to the customer's bonus balance per each 10 burles she paid.

Formally:

  1. a customer can choose any number x of bonuses to use (<image>)), 
  2. the customer's bonus balance is reduced by x, 
  3. the customer pays r - x burles, 
  4. the customer's bonus balance is increased by ⌊(r - x) / 10⌋ (i.e. integer division rounded down is used). 



Initially, there are b bonuses on Polycarp's account. Polycarp is going to have a lunch in "Ber Patio" for the next n days. He estimated the values a1, a2, ..., an, where ai is the number of burles in a receipt for the i-th day. The sum over all receipts doesn't exceed 105 burles.

Write a program to find the minimum number of burles Polycarp has to spend and an optimal strategy to use bonuses.

Input

The first line contains two integer numbers n and b (1 ≤ n ≤ 5000, 0 ≤ b ≤ 105) — number of days and initial number of bonuses Polycarp has.

The second line contains the integer sequence a1, a2, ..., an (1 ≤ ai ≤ 1000), where ai is the amount of burles in the i-th day's receipt.

It is guaranteed that the sum of all receipts does not exceed 105 burles.

Output

On the first line, print the expected minimal number of burles to pay for all n receipts.

On the second line, print the sequence of integer numbers b1, b2, ..., bn, where bi is the number of bonuses to use on the i-th day. If there are multiple solutions, print any of them.

Examples

Input

3 21
12 75 52


Output

110
2 5 22 


Input

3 39
58 64 33


Output

107
28 4 16 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int a[5500], use[5500];
int main() {
  int n, b;
  cin >> n >> b;
  for (int i = 1; i <= n; i++) cin >> a[i];
  int bonus = b;
  for (int i = 1; i <= n; i++) {
    use[i] = min(bonus, min(a[i] / 2, a[i] % 10));
    bonus -= use[i] - (a[i] - use[i]) / 10;
  }
  int lower_limit = 0;
  for (int i = n; i > 0; i--) {
    bonus += use[i] - (a[i] - use[i]) / 10;
    int reduced_cost =
        min(max(a[i] - bonus - 1, (a[i] - 1) / 2) / 10 * 10 + 10, a[i]);
    int used_bonus = a[i] - reduced_cost;
    if (used_bonus == 0) {
      lower_limit = max(use[i], lower_limit - (a[i] - use[i]) / 10 + use[i]);
      continue;
    }
    int rhs = -(bonus - used_bonus + (a[i] - used_bonus) / 10 - lower_limit);
    int real_used_bonus = used_bonus - max(0, (rhs + 10) / 11) * 10;
    use[i] = real_used_bonus;
    lower_limit = max(use[i], use[i] - (a[i] - use[i]) / 10 + lower_limit);
  }
  for (int j = 10; j > 0; j--) {
    bonus = b;
    for (int i = 1; i <= n; i++) bonus -= use[i] - (a[i] - use[i]) / 10;
    lower_limit = 0;
    for (int i = n; i > 0; i--) {
      bonus += use[i] - (a[i] - use[i]) / 10;
      int cur_lower_limit =
          max(use[i], use[i] - (a[i] - use[i]) / 10 + lower_limit);
      int max_added =
          min(a[i] / 2 - use[i], bonus - cur_lower_limit -
                                     ((a[i] - use[i]) % 10 == 0 &&
                                      lower_limit - (a[i] - use[i]) / 10 >= 0));
      if (max_added == j) {
        use[i] += j;
        cur_lower_limit =
            max(use[i], use[i] - (a[i] - use[i]) / 10 + lower_limit);
      }
      lower_limit = cur_lower_limit;
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) ans += a[i] - use[i];
  printf("%d\n", ans);
  for (int i = 1; i <= n; i++) printf("%d ", use[i]);
  return 0;
}