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>
int a[5005], res[5005], use[5005], n;
int main() {
  scanf("%d%d\n", &n, &res[0]);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) res[i] = res[i - 1] + a[i] / 10;
  int minres = res[n];
  for (int i = n; i >= 1; i--) {
    int x = std::min(a[i] % 10, a[i] / 2);
    minres = std::min(minres, res[i - 1]);
    use[i] = std::min(x, minres);
    minres -= use[i];
  }
  for (int i = 1; i <= n; i++)
    res[i] = res[i - 1] - use[i] + (a[i] - use[i]) / 10;
  minres = res[n];
  for (int i = n; i >= 1; i--) {
    int x =
        std::min((std::min(res[i - 1], a[i] / 2) - use[i]) / 10, minres / 11);
    use[i] += 10 * x;
    minres = std::min(minres - 11 * x, res[i - 1] - use[i]);
  }
  for (int cur = 9; cur; cur--) {
    for (int i = 1; i <= n; i++)
      res[i] = res[i - 1] - use[i] + (a[i] - use[i]) / 10;
    minres = res[n];
    for (int i = n; i >= 1; i--) {
      if (use[i] + cur <= std::min(res[i - 1], a[i] / 2) && minres >= cur + 1)
        use[i] += cur, minres -= cur + 1;
      minres = std::min(minres, res[i - 1] - use[i]);
    }
  }
  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]);
}