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 MAX = 1e4 + 5;
const int MAXN = 5000 + 5;
const int INF = 1e9;
short par[MAXN][MAX];
int a[MAXN], sum[MAXN];
int dp[2][MAX];
int main() {
  int n, b;
  cin >> n >> b;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sum[i] = sum[i - 1] + a[i];
  }
  int cur = 0;
  for (int i = 0; i < MAX; i++) dp[0][i] = dp[1][i] = INF;
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < MAX; j++) dp[1 - cur][j] = INF;
    for (int j = 0; j < MAX; j++) {
      if (dp[cur][j] == INF) continue;
      int cost = dp[cur][j];
      int s = sum[i - 1];
      int have = b + j - (s - cost);
      for (int k = 0; k <= min(a[i] / 2, have); k++) {
        int newCost = cost + a[i] - k;
        int newCreate = j + (a[i] - k) / 10;
        if (dp[1 - cur][newCreate] > newCost) {
          dp[1 - cur][newCreate] = newCost;
          par[i][newCreate] = a[i] - k;
        }
      }
    }
    cur = 1 - cur;
  }
  int ans = INF, create = -1;
  for (int i = 0; i < MAX; i++)
    if (dp[cur][i] < ans) {
      create = i;
      ans = dp[cur][i];
    }
  cout << ans << endl;
  vector<int> ans2;
  int i = n;
  while (i > 0) {
    ans2.push_back(par[i][create]);
    create = create - par[i][create] / 10;
    i--;
  }
  reverse(ans2.begin(), ans2.end());
  for (int i = 1; i <= ans2.size(); i++) cout << a[i] - ans2[i - 1] << " ";
  cout << endl;
  return 0;
}