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;
const int N(5555);
vector<int> f, lck[11];
int ans[N], a[N], p[N];
void spend_fre(int& tot) {
  while (tot && f.size()) {
    ans[f.back()]++;
    f.pop_back();
    tot--;
  }
}
void spend(int& tot) {
  spend_fre(tot);
  for (int i(10); i >= 1; i--) {
    while (tot && lck[i].size()) {
      tot--;
      f.insert(f.end(), i, lck[i].back());
      lck[i].pop_back();
      spend_fre(tot);
    }
  }
}
int main() {
  int n, b;
  scanf("%d%d", &n, &b);
  for (int i(1); i <= n; i++) {
    scanf("%d", &a[i]);
    p[i + 1] = a[i] / 10;
  }
  p[1] = b;
  for (int i(n); i >= 1; i--) {
    int lck_fre(p[i + 1]);
    int max_points(a[i] / 2);
    int fre(min(a[i] % 10, max_points));
    f.insert(f.end(), fre, i);
    max_points -= fre;
    for (int j(1); j <= max_points / 10; j++) {
      if (lck_fre) {
        lck_fre--;
        f.insert(f.end(), 10, i);
      } else {
        lck[10].push_back(i);
      }
    }
    max_points %= 10;
    if (max_points) {
      if (lck_fre) {
        f.insert(f.end(), max_points, i);
      } else {
        lck[max_points].push_back(i);
      }
    }
    spend(p[i]);
  }
  int tot(0);
  for (int i(1); i <= n; i++) {
    tot += a[i] - ans[i];
  }
  printf("%d\n", tot);
  for (int i(1); i <= n; i++) {
    printf("%d%c", ans[i], i == n ? '\n' : ' ');
  }
}