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, b;
int f[5002][10002];
short a[5002], trans[5002][10002], res[10002];
int main() {
  scanf("%d%d", &n, &b);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  f[0][0] = 0;
  int mx = 0, sa = 0;
  for (int i = 1; i <= n; i++) {
    int cur = a[i];
    fill(f[i], f[i] + mx + a[i] / 10 + 1, 0x3f3f3f3f);
    for (int j = 0; j <= mx; j++) {
      if (f[i - 1][j] != 0x3f3f3f3f) {
        int left = min(b - (sa - f[i - 1][j]) + j, a[i] / 2);
        for (int k = 0; k <= left; k++) {
          int cc = f[i - 1][j] + cur - k;
          int& s = f[i][j + (cur - k) / 10];
          if (s > cc) {
            s = cc;
            trans[i][j + (cur - k) / 10] = k;
          }
        }
      }
    }
    sa += a[i];
    mx += a[i] / 10;
  }
  int des = min_element(f[n], f[n] + mx + 1) - f[n];
  printf("%d\n", f[n][des]);
  for (int i = n; i >= 1; i--) {
    short v = trans[i][des];
    res[i] = v;
    des = des - (a[i] - v) / 10;
  }
  for (int i = 1; i <= n; i++) printf("%d ", res[i]);
  return 0;
}