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\nprint = stdout.write\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+1)st number in the output permutation at this i, and continue to move left. \n\n# segment tree and binary search\n\n_ = input()\nx = [int(i) for i in input().split()]\n \nres = []\n \nfrom math import log\n \n \nclass SegmentTree(object):\n \n    def __init__(self, nums):\n        self.arr = nums\n        self.l = len(nums)\n        self.tree = [0] * self.l + nums\n        for i in range(self.l - 1, 0, -1):\n            self.tree[i] = self.tree[i << 1] + self.tree[i << 1 | 1]\n \n    def update(self, i, val):\n        n = self.l + i\n        self.tree[n] = val\n        while n > 1:\n            self.tree[n >> 1] = self.tree[n] + self.tree[n ^ 1]\n            n >>= 1\n \n    def query(self, i, j):\n        m = self.l + i\n        n = self.l + j\n        res = 0\n        while m <= n:\n            if m & 1:\n                res += self.tree[m]\n                m += 1\n            m >>= 1\n            if n & 1 == 0:\n                res += self.tree[n]\n                n -= 1\n            n >>= 1\n        return res\n \ntree = SegmentTree(list(range(1, len(x) + 1)))\norg = len(x)\nwhile x:\n    # from back to forth\n    q = x.pop()\n \n    lo = 0\n    hi = org - 1\n \n    while lo < hi:\n        mid = (lo + hi) // 2\n        # print(lo, hi, mid)\n        sm = tree.query(0, mid)\n        # print(sm, mid)\n        if sm > q:\n            hi = mid\n        else:\n            lo = mid + 1\n    # print(tree.arr, lo, hi)\n    idx = tree.arr[lo]\n    # print(idx)\n    tree.update(lo, 0)\n    # also from back to forth\n    res.append(idx)\n \nprint(' '.join(str(i) for i in res[::-1]))"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13573