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 = 'import sys\ninput = sys.stdin.readline\n\nnn = 18\nbit=[0]*(2**nn+1)\n \ndef addbit(i, x):\n    while i <= 2**nn:\n        bit[i] += x\n        i += i & (-i)\n \ndef getsum(i):\n    ret = 0\n    while i != 0:\n        ret += bit[i]\n        i -= i&(-i)\n    return ret\n \ndef searchbit(x):\n    l, sl = 0, 0\n    d = 2**(nn-1)\n    while d:\n        m = l + d\n        sm = sl + bit[m]\n        if sm <= x:\n            l, sl = m, sm\n        d //= 2\n    return l + 1\n\nn = int(input())\nl = list(map(int, input().split()))\nfor i in range(1, n + 1):\n\taddbit(i, i)\nans = [0 for _ in range(n)]\nfor i in range(n - 1, -1, -1):\n\ta = searchbit(l[i])\n\taddbit(a, -a)\n\tans[i] = a\nprint(*ans)\n\n\n\n\n\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 2110961