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 a[5005], x[5005][12], y[5005][12];
int n, b, sa;
int min(int x, int y) { return x < y ? x : y; }
int fldiv(int x, int y) { return (x < 0 ? x - y + 1 : x) / y; }
int getp(int X, int Y) {
  if (Y < x[X][11] || Y < 0) return y[X][0];
  int d = 1;
  for (; x[X][d] > Y; d++)
    ;
  return y[X][d] + fldiv(Y - x[X][d], d);
}
int DP(int x, int y) {
  int ans = 1e9;
  for (int i = 0; i <= y && i * 2 <= a[x]; i++)
    ans = min(ans, getp(x + 1, y - i + (a[x] - i) / 10) - (a[x] - i) / 10);
  return ans;
}
int main() {
  scanf("%d%d", &n, &b);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]), sa += a[i];
  for (int i = n - 1; i >= 0; i--) {
    y[i][0] = DP(i, 0);
    int l = 0, r = 100000000;
    while (l < r) {
      int mid = (l + r) / 2;
      if (DP(i, mid) > y[i][0])
        r = mid;
      else
        l = mid + 1;
    }
    x[i][11] = l - 11;
    y[i][11] = y[i][0];
    for (int j = 10; j; j--) {
      int stx = x[i][j + 1], sty = y[i][j + 1];
      int l = 0, r = (1e9 - stx) / (j + 1);
      while (l < r) {
        int mid = (l + r + 1) / 2;
        if (DP(i, stx + mid * (j + 1) - 1) == sty + mid - 1)
          l = mid;
        else
          r = mid - 1;
      }
      x[i][j] = stx + l * (j + 1);
      y[i][j] = y[i][j + 1] + l;
    }
  }
  printf("%d\n", sa + getp(0, b) - b);
  for (int i = 0, j = b, r; i < n; j = j - r + (a[i++] - r) / 10) {
    r = 0;
    int now = getp(i, j);
    for (; now != getp(i + 1, j - r + (a[i] - r) / 10) - (a[i] - r) / 10;) r++;
    printf("%d ", r);
  }
}