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 = 5005;
const int MAXM = 100005;
int n, m;
int arr[MAXN], sum[MAXN];
int dp[2][10005], ans[MAXN][10005];
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    sum[i] = arr[i] + sum[i - 1];
  }
  memset(dp, 0x3f, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 0; i < n; i++) {
    memset(dp[(i + 1) & 1], 0x3f, sizeof(dp[(i + 1) & 1]));
    for (int j = 0; j < MAXM / 10; j++) {
      if (dp[i & 1][j] == 0x3f3f3f3f) continue;
      int left = m + j - (sum[i] - dp[i & 1][j]);
      for (int k = 0; k <= left && k <= arr[i + 1] / 2; k++) {
        int nj = j + (arr[i + 1] - k) / 10;
        if (dp[i & 1][j] + arr[i + 1] - k < dp[(i + 1) & 1][nj]) {
          dp[(i + 1) & 1][nj] = dp[i & 1][j] + arr[i + 1] - k;
          ans[i + 1][nj] = k;
        }
      }
    }
  }
  int best = 0x3f3f3f3f;
  int idx;
  for (int i = 0; i < MAXM / 10; i++) {
    if (dp[n & 1][i] < best) {
      best = dp[n & 1][i];
      idx = i;
    }
  }
  stack<int> s;
  cout << best << endl;
  for (int i = n; i >= 1; i--) {
    s.push(ans[i][idx]);
    idx = idx - (arr[i] - ans[i][idx]) / 10;
  }
  while (!s.empty()) {
    cout << s.top() << " \n"[s.size() == 1];
    s.pop();
  }
  return 0;
}