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>
const int inf = 1000000000, m = 10000;
int p[5005][10005], f[10005], g[10005], a[5005], sum[5005], use[5005], n, B;
int main() {
  scanf("%d%d\n", &n, &B);
  for (int i = 1; i <= m; i++) f[i] = g[i] = inf;
  int cur = 0;
  g[0] = 0;
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    a[i] = x;
    sum[i] = sum[i - 1] + x;
    cur = std::min(cur + x / 10, m);
    for (int j = 0; j <= cur; j++) f[j] = g[j], g[j] = inf;
    for (int j = 0; j <= cur; j++)
      if (f[j] <= sum[i - 1]) {
        int b = B - sum[i - 1] + f[j] + j;
        for (int k = 0; k <= x / 2 && k <= b; k++) {
          int y = j + (x - k) / 10;
          if (f[j] + x - k < g[y]) {
            g[y] = f[j] + x - k;
            p[i][y] = k;
          }
        }
      }
  }
  int ans = 0;
  for (int i = 1; i <= cur; i++)
    if (g[i] < g[ans]) ans = i;
  printf("%d\n", g[ans]);
  for (int i = n; i >= 1; ans -= (a[i] - p[i][ans]) / 10, i--)
    use[i] = p[i][ans];
  for (int i = 1; i <= n; i++) printf("%d ", use[i]);
}