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 = "\n\ndef sum_number(n,j):\n    j[0]=0\n    j[1]=0\n    for i in range(2,n+1):\n        j[i]=j[i-1]+(i-1)\n    return(j)\npo=int(input())\nl=[0]*(po+1)\nl1=[int(i) for i in input().split()]\ndef getsum(BITTree,i):\n    s = 0\n    while i > 0:\n        s += BITTree[i]\n        i -= i & (-i) \n    return(s) \ndef updatebit(BITTree , n , i ,v):\n    #print('n',n)\n    while i <= n:\n        #print('i',i)\n        BITTree[i] += v\n        i += i & (-i)\n    #print(BITTree)\nfor i in range(1,po+1):\n    updatebit(l,po,i,i)\noutput=[0]*po\nfor i in range(po-1,-1,-1):\n    min_=0\n    max_=po\n    k=l1[i]\n    while True:\n        x=(min_+max_+1)//2\n        if getsum(l,x)>k:\n            if getsum(l,x-1)==k:\n                output[i]=x\n                break\n            else:\n                #print(x)\n                max_=x\n        else :\n            #print(x)\n            min_=x\n    updatebit(l,po,x,-x)\nprint(*output)\n\n\n\n        \n    \n    \n\n\n\n"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13329