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 mn = 5005;
int n, b, sb;
int a[mn];
int d[2][mn * 2];
short fa[mn][mn * 2];
short used[mn][mn * 2];
int sa[mn];
int main() {
  scanf("%d%d", &n, &b);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), sa[i] = sa[i - 1] + a[i];
  bool p = 0;
  for (int i = 0; i < n; i++) {
    memset(d[p ^ 1], 1, sizeof d[p ^ 1]);
    for (int j = 0; j <= sb; j++) {
      if (d[p][j] > 1e7) continue;
      int nowb = b + j - (sa[i] - d[p][j]);
      int t = min({a[i + 1] / 2, nowb});
      for (int k = 0; k <= t; k++) {
        int v = j + (a[i + 1] - k) / 10;
        if (d[p ^ 1][v] > d[p][j] + a[i + 1] - k) {
          fa[i + 1][v] = j;
          used[i + 1][v] = k;
          d[p ^ 1][v] = d[p][j] + a[i + 1] - k;
        }
      }
    }
    sb += a[i + 1] / 10;
    p ^= 1;
  }
  int ans = 1e5 + 5, choose = 0;
  for (int j = 0; j <= sb; j++) {
    if (ans > d[p][j]) {
      ans = d[p][j];
      choose = j;
    }
  }
  printf("%d\n", ans);
  vector<int> v;
  for (int i = n; i; i--) {
    v.push_back(used[i][choose]);
    choose = fa[i][choose];
  }
  reverse(v.begin(), v.end());
  for (int i = 1; i <= n; i++) {
    printf("%d ", v[i - 1]);
  }
  return 0;
}