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 n, coin, money;
int a[5555];
int expire[5555];
vector<int> fre, locked[11];
int ans[5555];
void add_fre(int amount, int id) {
  for (int i = 0; i < amount; i++) fre.push_back(id);
}
void try_spend_fre(int &amount) {
  while (amount && fre.size()) {
    ans[fre.back()]++;
    amount--;
    fre.pop_back();
  }
}
void try_spend(int &amount) {
  try_spend_fre(amount);
  for (int i = 10; i > 0 && amount; i--)
    while (amount && locked[i].size()) {
      amount--;
      int id = locked[i].back();
      locked[i].pop_back();
      add_fre(i, id);
      try_spend_fre(amount);
    }
}
int main() {
  scanf("%d%d", &n, &coin);
  money = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    int today_coin = a[i] / 10;
    expire[i + 1] = today_coin;
    money += a[i];
  }
  for (int i = n; i > 0; i--) {
    int today_discount = a[i] / 2;
    int lock_fre = expire[i + 1];
    int today_fre = min(a[i] % 10, today_discount);
    add_fre(today_fre, i);
    today_discount -= today_fre;
    for (int j = today_discount / 10; j > 0; j--)
      if (lock_fre) {
        lock_fre--;
        add_fre(10, i);
      } else
        locked[10].push_back(i);
    today_discount %= 10;
    if (today_discount) {
      if (lock_fre)
        add_fre(today_discount, i);
      else
        locked[today_discount].push_back(i);
    }
    try_spend(expire[i]);
  }
  try_spend(coin);
  for (int i = 1; i <= n; i++) money -= ans[i];
  printf("%d\n", money);
  for (int i = 1; i <= n; i++) printf("%d%c", ans[i], i < n ? ' ' : '\n');
  return 0;
}