Problem:
Score : 700 points

Problem Statement

Snuke loves constructing integer sequences.

There are N piles of stones, numbered 1 through N.
The pile numbered i consists of a_i stones.

Snuke will construct an integer sequence s of length Σa_i, as follows:

Among the piles with the largest number of stones remaining, let x be the index of the pile with the smallest index. Append x to the end of s.

Select a pile with one or more stones remaining, and remove a stone from that pile.

If there is a pile with one or more stones remaining, go back to step 1. Otherwise, terminate the process.

We are interested in the lexicographically smallest sequence that can be constructed. For each of the integers 1,2,3,...,N, how many times does it occur in the lexicographically smallest sequence?

Constraints

1 ≤ N ≤ 10^{5}

1 ≤ a_i ≤ 10^{9}

Input

The input is given from Standard Input in the following format:

N
a_1 a_2 ... a_{N}

Output

Print N lines. The i-th line should contain the number of the occurrences of the integer i in the lexicographically smallest sequence that can be constructed.

Sample Input 1

2
1 2

Sample Output 1

2
1

The lexicographically smallest sequence is constructed as follows:

Since the pile with the largest number of stones remaining is pile 2, append 2 to the end of s. Then, remove a stone from pile 2.

Since the piles with the largest number of stones remaining are pile 1 and 2, append 1 to the end of s (we take the smallest index). Then, remove a stone from pile 2.

Since the pile with the largest number of stones remaining is pile 1, append 1 to the end of s. Then, remove a stone from pile 1.

The resulting sequence is (2,1,1). In this sequence, 1 occurs twice, and 2 occurs once.

Sample Input 2

10
1 2 1 3 2 4 2 5 8 1

Sample Output 2

10
7
0
4
0
3
0
2
3
0

Solution in Scala (2.11.7) (judge verdict: Wrong Answer)

import java.awt.geom.Point2D
import java.io._
import java.util.StringTokenizer

import scala.collection.mutable
import scala.util.Sorting
import math.{abs, max, min}
import scala.collection.mutable.{ArrayBuffer, ListBuffer}
import scala.reflect.ClassTag

object Main {
  val MOD = 1000000007
  val out = new PrintWriter(System.out)

  case class Barrier(x: Int, y: Int, r: Int)
  type WUGraph = Array[Barrier]
  case class Coordinate(x: Int, y: Int)

  def solve(): Unit = {
    val sc = new InputReader(System.in)
    val N = sc.nextInt()
    val A = map(N)(_ => sc.nextInt())

    val mxs: ArrayBuffer[Int] = {
      val mxs = ArrayBuffer.empty[Int]
      var mx = 0
      rep(N){ i =>
        if (A(i) > mx) {
          mx = A(i)
          mxs += i
        }
      }
      mxs
    }

    val inRanges = {
      val inRanges = Array.fill(mxs.length)(ArrayBuffer.empty[Int])
      val pq = new java.util.PriorityQueue[Int]
      val mxFlag = Array.ofDim[Boolean](N)
      mxs foreach (i => mxFlag(i) = true)
      rep(N)(i => if (!mxFlag(i)) pq.add(A(i)))
      rep(mxs.length) { i =>
        val mx = A(mxs(i))
        while(!pq.isEmpty && pq.peek() <= mx) {
          inRanges(i) += pq.poll()
        }
      }
      inRanges
    }

    val ans = Array.ofDim[Long](N)
    def step(i: Int, cnt: Int): Unit = {
      val inRange = inRanges(i)
      val mx = A(mxs(i))
      i match {
        case 0 =>
          ans(mxs(i)) = mx.toLong * (cnt + 1) + inRange.sum

        case _ =>
          val nextMx = A(mxs(i - 1))
          ans(mxs(i)) = (mx - nextMx).toLong * (cnt + 1) + inRange.sumBy(_ - nextMx)
          val nextCnt = cnt + 1 + inRange.length
          step(i - 1, nextCnt)
      }
    }
    step(mxs.length - 1, 0)

    rep(N)(i => out.println(ans(i)))
  }


  def main(args: Array[String]): Unit = {
    solve()
    out.flush()
  }


  class InputReader(val stream: InputStream) {
    private val reader = new BufferedReader(new InputStreamReader(stream), 32768)
    private var tokenizer: StringTokenizer = null

    def next(): String = {
      while (tokenizer == null || !tokenizer.hasMoreTokens)
        tokenizer = new StringTokenizer(reader.readLine)
      tokenizer.nextToken
    }

    def nextInt(): Int = next().toInt
    def nextLong(): Long = next().toLong
    def nextChar(): Char = next().charAt(0)
  }
  def rep(n: Int)(f: Int => Unit): Unit = {
    var i = 0
    while(i < n) { f(i); i += 1 }
  }
  def rep_r(n: Int)(f: Int => Unit): Unit = {
    var i = n - 1
    while(i >= 0) { f(i); i -= 1 }
  }

  def map[@specialized A: ClassTag](n: Int)(f: Int => A): Array[A] = {
    val res = Array.ofDim[A](n)
    rep(n)(i => res(i) = f(i))
    res
  }

  implicit class ArrayOpts[A](val as: Array[A]) extends AnyVal {
    // todo Orderingだとboxing発生するので自作Orderを用意したい
    def maxByOpt[B: Ordering](f: A => B): Option[A] = {
      if (as.nonEmpty) Some(as.maxBy(f)) else None
    }

    def grpBy[K](f: A => K): mutable.Map[K, ArrayBuffer[A]] = {
      val map = mutable.Map.empty[K, ArrayBuffer[A]]
      rep(as.length)(i => map.getOrElseUpdate(f(as(i)), ArrayBuffer()) += as(i))
      map
    }

    def sumBy[B](f: A => B)(implicit num: Numeric[B]): B = {
      var sum = num.zero
      rep(as.length)(i => sum = num.plus(sum, f(as(i))))
      sum
    }

    def minByEx[B](f: A => B, ixRange: Range = as.indices)(implicit cmp: Ordering[B]): (A, B) = {
      limit(f, ixRange)(cmp.lt)
    }

    def maxByEx[B](f: A => B, ixRange: Range = as.indices)(implicit cmp: Ordering[B]): (A, B) = {
      limit(f, ixRange)(cmp.gt)
    }

    private def limit[B](f: A => B, ixRange: Range)(cmp: (B, B) => Boolean): (A, B) = {
      var limA = as(ixRange.head)
      var limB = f(limA)

      for (i <- ixRange.tail) {
        val a = as(i)
        val b = f(a)
        if (cmp(b, limB)) {
          limA = a
          limB = b
        }
      }
      (limA, limB)
    }
  }

  implicit class IterableOpts[A](val as: Iterable[A]) extends AnyVal {
    def sumBy[B](f: A => B)(implicit num: Numeric[B]): B = {
      as.foldLeft(num.zero)((acc, a) => num.plus(acc, f(a)))
    }
  }
}