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 maxn = 10000;
const int inf = 1e9;
int n, b, s[5050], f[2][maxn + 5];
short pre[5050][maxn + 5];
int out[5050], a[5050];
int main() {
  scanf("%d%d", &n, &b);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), s[i] = s[i - 1] + a[i];
  for (int j = 0; j <= maxn; j++) f[0][j] = f[1][j] = inf;
  f[0][0] = 0;
  int op = 1;
  for (int i = 1; i <= n; i++) {
    op ^= 1;
    for (int j = 0; j <= maxn; j++) f[op ^ 1][j] = inf;
    for (int j = 0; j <= maxn; j++)
      if (f[op][j] != inf) {
        int tot = s[i - 1];
        int rest = b + j - (tot - f[op][j]);
        for (int k = 0; k <= min(rest, a[i] / 2); k++) {
          int now = f[op][j] + a[i] - k;
          int nj = j + (a[i] - k) / 10;
          if (now < f[op ^ 1][nj]) {
            f[op ^ 1][nj] = now;
            pre[i][nj] = a[i] - k;
          }
        }
      }
  }
  op ^= 1;
  int ans = inf, now = 0;
  for (int i = 0; i <= maxn; i++)
    if (ans > f[op][i]) ans = f[op][i], now = i;
  printf("%d\n", ans);
  for (int i = n; i >= 1; i--) {
    out[i] = pre[i][now];
    now = now - out[i] / 10;
    op ^= 1;
  }
  for (int i = 1; i <= n; i++) printf("%d ", a[i] - out[i]);
  return 0;
}