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 = 10005;
int dp[2][maxn];
int a[maxn];
const int inf = 1000000000;
int pre[5006][maxn];
int con[maxn];
int main() {
  int n, b;
  int u = 0;
  cin >> n >> b;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < (maxn); i++) dp[0][i] = dp[1][i] = inf;
  dp[0][0] = 0;
  int all = 0;
  for (int i = 1; i <= n; i++) {
    u = u ^ 1;
    for (int j = 0; j < maxn; j++) {
      if (dp[u ^ 1][j] != inf) {
        int rest = -(all - dp[u ^ 1][j]) + b + j;
        for (int f = 0; f <= min(rest, a[i] / 2); f++) {
          if (dp[u][j + (a[i] - f) / 10] > dp[u ^ 1][j] + a[i] - f) {
            dp[u][j + (a[i] - f) / 10] = dp[u ^ 1][j] + a[i] - f;
            pre[i][j + (a[i] - f) / 10] = f;
          }
        }
      }
    }
    for (int j = 0; j < maxn; j++) {
      dp[u ^ 1][j] = inf;
    }
    all += a[i];
  }
  int ans = inf;
  int rest = 0;
  for (int i = 0; i < maxn; i++) {
    if (ans > dp[u][i]) {
      ans = dp[u][i];
      rest = i;
    }
  }
  printf("%d\n", ans);
  for (int i = n; i >= 1; i--) {
    con[i] = pre[i][rest];
    rest = rest - (a[i] - pre[i][rest]) / 10;
  }
  for (int i = 1; i <= n; i++) printf("%d ", con[i]);
}