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 
#include <bits/stdc++.h>
const int N = 5054, M = 10054;
int n, B;
int a[N], A[N], ans[N];
int dp[2][M], *cur = *dp, *nxt = dp[1];
short from[N][M];
inline bool down(int &x, const int y) { return x > y ? x = y, 1 : 0; }
int main() {
  int i, j, nj, u, v, used, s = 0;
  scanf("%d%d", &n, &B);
  memset(nxt, 63, sizeof *dp), *nxt = 0;
  for (i = 1; i <= n; ++i) {
    scanf("%d", a + i), A[i] = A[i - 1] + a[i], s += a[i] / 10;
    std::swap(cur, nxt), memset(nxt, 63, sizeof *dp);
    for (j = 0; j <= s; ++j)
      if ((used = A[i - 1] - cur[j]) >= 0)
        for (u = 0, v = a[i]; u <= v && u <= B - used + j; ++u, --v)
          if (down(nxt[nj = j + v / 10], cur[j] + v)) from[i][nj] = u;
  }
  j = std::min_element(nxt, nxt + (s + 1)) - nxt;
  printf("%d\n", nxt[j]);
  for (i = n; i; --i) ans[i] = u = from[i][j], v = a[i] - u, j -= v / 10;
  for (i = 1; i <= n; ++i) printf("%d%c", ans[i], i == n ? 10 : 32);
  return 0;
}
