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-----
2
1 2

-----Sample Output-----
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.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = "# -*- coding: utf-8 -*-\n\nimport sys\nfrom collections import Counter\n\ndef input(): return sys.stdin.readline().strip()\ndef list2d(a, b, c): return [[c] * b for i in range(a)]\ndef list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]\ndef list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]\ndef ceil(x, y=1): return int(-(-x // y))\ndef INT(): return int(input())\ndef MAP(): return map(int, input().split())\ndef LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]\ndef Yes(): print('Yes')\ndef No(): print('No')\ndef YES(): print('YES')\ndef NO(): print('NO')\nsys.setrecursionlimit(10 ** 9)\nINF = 10 ** 18\nMOD = 10 ** 9 + 7\n\nN = INT()\nA = LIST()\n\nC = Counter(A)\n# 重複削除して降順、この順に処理していく\nunique = sorted(set(A), reverse=1) + [0]\n# ある値が最初に出現するindexを保持\nD = {}\nfor i in range(N-1, -1, -1):\n    D[A[i]] = i\n\nans = [0] * N\nprevcnt = 0\nmnidx = INF\nfor i, a in enumerate(unique[:-1]):\n    nxta = unique[i+1]\n    # 今回の位置が選ばれる回数 = 次の値との差分 * 今の値の個数\n    cnt = (a - nxta) * (C[a] + prevcnt)\n    # 今の値の先頭index\n    mnidx = min(mnidx, D[a])\n    ans[mnidx] += cnt\n    # 今回の個数を次回以降用に追加\n    prevcnt += C[a]\n[print(a) for a in ans]\n"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 16181