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 = 10005;
template <typename T>
inline void read(T &AKNOI) {
  T x = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  AKNOI = flag * x;
}
int n, b, a[MAXN];
int dp[2][MAXM], ds[MAXN][MAXM], ans[MAXN];
void solve() {
  read(n);
  read(b);
  memset(dp[0], 0x3f, sizeof(dp[0]));
  dp[0][0] = 0;
  int pre = 0, cur = 1, up = 0;
  for (int i = 1, x, sum = 0; i <= n; ++i) {
    read(x);
    a[i] = x;
    memset(dp[cur], 0x3f, sizeof(dp[cur]));
    for (int j = 0; j <= up; ++j) {
      int p = dp[pre][j];
      if (p > sum) continue;
      int left = b - (sum - p) + j;
      for (int k = 0; k <= left && k <= (x >> 1); ++k) {
        int t = j + (x - k) / 10;
        if (p + x - k < dp[cur][t]) {
          dp[cur][t] = p + x - k;
          ds[i][t] = k;
        }
      }
    }
    sum += x;
    up += x / 10;
    swap(pre, cur);
  }
  int ansp = 0;
  for (int i = 1; i <= up; ++i) {
    if (dp[pre][i] < dp[pre][ansp]) {
      ansp = i;
    }
  }
  printf("%d\n", dp[pre][ansp]);
  for (int i = n; i >= 1; --i) {
    ans[i] = ds[i][ansp];
    ansp -= (a[i] - ds[i][ansp]) / 10;
  }
  for (int i = 1; i <= n; ++i) {
    printf("%d ", ans[i]);
  }
}
int main() {
  solve();
  return 0;
}