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=int(input())\nP=[int(i) for i in input().split()]\nimport math\nimport bisect\n\nn_max=2*10**5\nnn=int(math.log2(n_max))+1\nBIT=[0]*(2**nn+1)\n\ndef bitsum(BIT,i):\n  s=0\n  while i>0:\n    s+=BIT[i]\n    i-=i&(-i)\n  return s\ndef bitadd(BIT,i,x):\n  if i<=0:\n    return True\n  else:\n    while i<=2**nn:\n      BIT[i]+=x\n      i+=i&(-i)\n    return BIT\ndef bitlowerbound(BIT,s):\n  if s<=0:\n    return 0\n  else:\n    ret=0\n    k=2**nn\n    while k>0:\n      if k+ret<=2**nn and BIT[k+ret]<s:\n        s-=BIT[k+ret]\n        ret+=k\n      k//=2\n    return ret+1\nfor i in range(n_max):\n  bitadd(BIT,i+1,i+1)\n\nAns=[]\nfor i in reversed(range(n)):\n  p=P[i]\n  ans=bitlowerbound(BIT,p+1)\n  Ans.append(ans)\n  bitadd(BIT,ans,-ans)\nAns=Ans[::-1]\nprint(*Ans)'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 8498879