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 Lua (5.3.2) (judge verdict: Accepted)

local function comp(a, b)
  return a < b
end

local function lower_bound(ary, x)
  local num = #ary
  if num == 0 then return 1 end
  if not comp(ary[1], x) then return 1 end
  if comp(ary[num], x) then return num + 1 end
  local min, max = 1, num
  while 1 < max - min do
    local mid = (min + max) // 2
    if comp(ary[mid], x) then
      min = mid
    else
      max = mid
    end
  end
  return max
end

local n = io.read("*n")
local stage = {1}
local stagenum = {io.read("*n")}
local stagesum = {stagenum[1]}
local stageaddcnt = {1}
for i = 2, n do
  local a = io.read("*n")
  local pos = lower_bound(stagenum, a)
  if #stagenum < pos then
    table.insert(stage, i)
    table.insert(stagenum, a)
    table.insert(stagesum, a)
    table.insert(stageaddcnt, 1)
  else
    stagesum[pos] = stagesum[pos] + a
    stageaddcnt[pos] = stageaddcnt[pos] + 1
  end
end
local ret = {}
for i = 1, n do
  ret[i] = 0
end

for i = #stagesum, 2, -1 do
  local dst_i = stage[i]
  local num = stagenum[i]
  local sum = stagesum[i]
  local addcnt = stageaddcnt[i]
  ret[dst_i] = sum - addcnt * stagenum[i - 1]
  stageaddcnt[i - 1] = stageaddcnt[i - 1] + addcnt
  stagesum[i - 1] = stagesum[i- 1] + stagenum[i - 1] * addcnt
end
ret[1] = stagesum[1]

print(table.concat(ret, "\n"))