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 N = 5005, M = 10005;
int n, m, sum, now;
int dp[2][M], a[N], f[N][M], Ans[N];
int main() {
  while (~scanf("%d%d", &n, &m)) {
    sum = 0, now = 0;
    memset(dp, 0x3f, sizeof(dp));
    dp[now][0] = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      now = now ^ 1;
      for (int j = 0; j <= sum / 10; j++) dp[now][j] = 0x3f3f3f3f;
      for (int j = 0; j <= sum / 10; j++) {
        int rest = m - (sum - dp[now ^ 1][j]) + j;
        for (int k = 0; k <= std::min(a[i] / 2, rest); k++) {
          int cst = a[i] - k;
          if (dp[now ^ 1][j] + cst < dp[now][j + cst / 10]) {
            dp[now][j + cst / 10] = dp[now ^ 1][j] + cst;
            f[i][j + cst / 10] = k;
          }
        }
      }
      sum += a[i];
    }
    int ans = sum, tot;
    for (int i = 0; i <= sum / 10; i++)
      if (dp[now][i] < ans) ans = dp[now][i], tot = i;
    printf("%d\n", ans);
    for (int i = n; i >= 1; i--)
      Ans[i] = f[i][tot], tot -= (a[i] - f[i][tot]) / 10;
    for (int i = 1; i <= n; i++) printf("%d%c", Ans[i], (i == n ? '\n' : ' '));
  }
}