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 = 5050, Maxstate = 10020, Inf = 1e9;
int n, b;
int a[Maxn], sum[Maxn];
int dp[2][10020];
short pre[Maxn][Maxstate];
void init(int cs) {
  for (int i = 0; i < Maxstate; i++) dp[cs][i] = Inf;
}
int main() {
  scanf("%d%d", &n, &b);
  for (int i = 1; i <= n; i++) scanf("%d", a + i), sum[i] = sum[i - 1] + a[i];
  int cs = 0;
  init(0);
  init(1);
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++, cs ^= 1) {
    init(cs ^ 1);
    for (int j = 0; j < Maxstate; j++) {
      if (dp[cs][j] == Inf) continue;
      int w = dp[cs][j];
      int total = sum[i - 1];
      int rest = b + j - (total - w);
      for (int k = 0; k <= rest && k <= a[i] / 2; k++) {
        int nw = w + a[i] - k;
        int nj = j + (a[i] - k) / 10;
        if (dp[cs ^ 1][nj] > nw) {
          dp[cs ^ 1][nj] = nw;
          pre[i][nj] = a[i] - k;
        }
      }
    }
  }
  int ans = Inf, cur = -1;
  for (int i = 0; i < Maxstate; i++)
    if (dp[cs][i] < ans) {
      cur = i;
      ans = dp[cs][i];
    }
  printf("%d\n", ans);
  vector<int> rep;
  for (int i = n; i > 0; i--) {
    rep.push_back(pre[i][cur]);
    cur = cur - pre[i][cur] / 10;
  }
  reverse(rep.begin(), rep.end());
  for (int i = 0; i < rep.size(); i++)
    printf("%d%c", a[i + 1] - rep[i], i == rep.size() - 1 ? '\n' : ' ');
  return 0;
}