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 main() {
  int n, b;
  cin >> n >> b;
  vector<int> cost(n), lim(n), bonus(n);
  for (int i = 0; i < n; i++) {
    cin >> cost[i];
    lim[i] = cost[i] / 2;
  }
  auto valid = [&]() {
    int x = b;
    for (int i = 0; i < n; i++) {
      if (bonus[i] > x) {
        return false;
      }
      if (bonus[i] > lim[i]) {
        return false;
      }
      x -= bonus[i];
      x += cost[i] / 10;
    }
    return true;
  };
  int arr[] = {1, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1};
  for (int z = 0; z < sizeof(arr) / sizeof(arr[0]); z++) {
    int amount = arr[z];
    for (int i = n - 1; i >= 0; i--) {
      while (true) {
        if (z == 0 && cost[i] % 10 == 0) {
          break;
        }
        bonus[i] += amount;
        cost[i] -= amount;
        if (!valid()) {
          bonus[i] -= amount;
          cost[i] += amount;
          break;
        }
      }
    }
  }
  int tot = 0;
  for (int i = 0; i < n; i++) {
    tot += cost[i];
  }
  cout << tot << endl;
  for (int i = 0; i < n; i++) {
    if (i > 0) cout << ' ';
    cout << bonus[i];
  }
  cout << endl;
}