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;
const int INF = 0x3f3f3f3f;
short fa[MAXN][MAXM];
int dp[2][MAXM];
int a[MAXN];
int n, m;
void solve() {
  memset(dp, INF, sizeof dp);
  memset(fa, 0, sizeof fa);
  dp[0][0] = 0;
  int sum = 0, o = 1;
  for (int i = 1; i <= n; ++i, o ^= 1) {
    scanf("%d", &a[i]);
    int M = (sum + a[i]) / 10;
    for (int j = 0; j <= M; ++j) {
      dp[o][j] = INF;
    }
    for (int j = 0; j <= M; ++j)
      if (dp[o ^ 1][j] != INF) {
        int w = dp[o ^ 1][j];
        int has = m + j - (sum - w);
        int t = min(has, a[i] / 2);
        for (int k = 0; k <= t; ++k) {
          int nxtw = w + a[i] - k;
          int nxtj = j + (a[i] - k) / 10;
          if (dp[o][nxtj] > nxtw) {
            dp[o][nxtj] = nxtw;
            fa[i][nxtj] = a[i] - k;
          }
        }
      }
    sum += a[i];
  }
  int ans = INF, pos;
  for (int i = 0; i < MAXM; ++i) {
    if (ans > dp[o ^ 1][i]) {
      ans = dp[o ^ 1][i];
      pos = i;
    }
  }
  vector<int> res;
  for (int i = n; i >= 1; --i) {
    res.push_back(fa[i][pos]);
    pos -= fa[i][pos] / 10;
  }
  reverse(res.begin(), res.end());
  printf("%d\n", ans);
  for (int i = 0; i < n; ++i) {
    printf("%d%c", a[i + 1] - res[i], i < n - 1 ? ' ' : '\n');
  }
}
int main() {
  while (~scanf("%d%d", &n, &m)) solve();
  return 0;
}