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 nax = 5005;
int cost[nax];
int would[nax];
int use[nax];
void calc_would(int n) {
  for (int i = 1; i <= n; ++i)
    would[i] = would[i - 1] - use[i] + (cost[i] - use[i]) / 10;
}
int main(void) {
  int n;
  scanf("%d%d", &n, &would[0]);
  for (int i = 1; i <= n; ++i) scanf("%d", &cost[i]);
  calc_would(n);
  int tmp = would[n];
  for (int i = n; i >= 1; --i) {
    int czapa = min(cost[i] / 2, cost[i] % 10);
    tmp = min(tmp, would[i - 1]);
    use[i] = min(czapa, tmp);
    tmp -= use[i];
  }
  for (int cnt = 10; cnt >= 1; --cnt) {
    calc_would(n);
    tmp = would[n];
    for (int i = n; i >= 1; --i) {
      while (use[i] + cnt <= cost[i] / 2) {
        if (would[i - 1] - use[i] >= cnt && tmp >= cnt + 1) {
          use[i] += cnt;
          tmp -= cnt + 1;
        } else
          break;
      }
      tmp = min(tmp, would[i - 1] - use[i]);
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i) ans += cost[i] - use[i];
  printf("%d\n", ans);
  for (int i = 1; i <= n; ++i) printf("%d ", use[i]);
  puts("");
  return 0;
}