An array of integers $p_{1},p_{2}, \ldots,p_{n}$ is called a permutation if it contains each number from $1$ to $n$ exactly once. For example, the following arrays are permutations: $[3,1,2], [1], [1,2,3,4,5]$ and $[4,3,1,2]$. The following arrays are not permutations: $[2], [1,1], [2,3,4]$.

There is a hidden permutation of length $n$.

For each index $i$, you are given $s_{i}$, which equals to the sum of all $p_{j}$ such that $j < i$ and $p_{j} < p_{i}$. In other words, $s_i$ is the sum of elements before the $i$-th element that are smaller than the $i$-th element.

Your task is to restore the permutation.


-----Input-----

The first line contains a single integer $n$ ($1 \le n \le 2 \cdot 10^{5}$) — the size of the permutation.

The second line contains $n$ integers $s_{1}, s_{2}, \ldots, s_{n}$ ($0 \le s_{i} \le \frac{n(n-1)}{2}$).

It is guaranteed that the array $s$ corresponds to a valid permutation of length $n$.


-----Output-----

Print $n$ integers $p_{1}, p_{2}, \ldots, p_{n}$ — the elements of the restored permutation. We can show that the answer is always unique.


-----Examples-----
Input
3
0 0 0

Output
3 2 1

Input
2
0 1

Output
1 2

Input
5
0 1 1 1 10

Output
1 4 3 2 5



-----Note-----

In the first example for each $i$ there is no index $j$ satisfying both conditions, hence $s_i$ are always $0$.

In the second example for $i = 2$ it happens that $j = 1$ satisfies the conditions, so $s_2 = p_1$.

In the third example for $i = 2, 3, 4$ only $j = 1$ satisfies the conditions, so $s_2 = s_3 = s_4 = 1$. For $i = 5$ all $j = 1, 2, 3, 4$ are possible, so $s_5 = p_1 + p_2 + p_3 + p_4 = 10$.

```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 = "# https://codeforces.com/contest/1208/problem/D\n\nfrom sys import stdin, stdout\ninput = stdin.readline\n# print = stdout.write\n\n# si is the sum of elements before the i-th element that are smaller than the i-th element.\n\n# For every i from N to 1, let's say the value of the si is x.\n# So it means there are k smallest unused numbers whose sum is x.\n# We simply put the k+1st number in the output permutation at this i, and continue to move left.\n\n# BIT and binary lifting\n# https://codeforces.com/contest/1208/submission/59526098\n\n\nclass BIT:\n    def __init__(self, nums):\n        # we store the sum information in bit 1.\n        # so the indices should be 1 based.\n        # here we assume nums[0] = 0\n        self.nums = [0] * (len(nums))\n        for i, x in enumerate(nums):\n            if i == 0:\n                continue\n            self.update(i, x)\n\n    def low_bit(self, x):\n        return x & (-x)\n\n    def update(self, i, diff):\n        while i < len(self.nums):\n            self.nums[i] += diff\n            i += self.low_bit(i)\n\n    def prefix_sum(self, i):\n        ret = 0\n        while i != 0:\n            ret += self.nums[i]\n            i -= self.low_bit(i)\n        return ret\n\n    def search(self, x):\n        # find the index i such that prefix_sum(i) == x\n        cur_index, cur_sum = 0, 0\n        delta = len(self.nums) - 1\n        while delta - self.low_bit(delta):\n            delta -= self.low_bit(delta)\n\n        while delta:            \n            m = cur_index + delta\n            if m < len(self.nums):\n                sm = cur_sum + self.nums[m]\n                if sm <= x:\n                    cur_index, cur_sum = m, sm\n            delta //= 2\n        return cur_index + 1\n\n\nn = int(input())\nbit = BIT(list(range(n+1)))\n\nans = [0 for _ in range(n)]\nnums = list(map(int, input().split()))\nfor i in range(n - 1, -1, -1):\n    index = bit.search(nums[i])\n    bit.update(index, -index)\n    ans[i] = index\nprint(*ans)\n"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 16615