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 $s_5 = p_1 + p_2 + p_3 + p_4 = 10$.
n=int(input())
P=[int(i) for i in input().split()]
import math
import bisect

n_max=2*10**5
nn=int(math.log2(n_max))+1
BIT=[0]*(2**nn+1)

def bitsum(BIT,i):
  s=0
  while i>0:
    s+=BIT[i]
    i-=i&(-i)
  return s
def bitadd(BIT,i,x):
  if i<=0:
    return True
  else:
    while i<=2**nn:
      BIT[i]+=x
      i+=i&(-i)
    return BIT
def bitlowerbound(BIT,s):
  if s<=0:
    return 0
  else:
    ret=0
    k=2**nn
    while k>0:
      if k+ret<=2**nn and BIT[k+ret]<s:
        s-=BIT[k+ret]
        ret+=k
      k//=2
    return ret+1
for i in range(n_max):
  bitadd(BIT,i+1,i+1)

Ans=[]
for i in reversed(range(n)):
  p=P[i]
  ans=bitlowerbound(BIT,p+1)
  Ans.append(ans)
  bitadd(BIT,ans,-ans)
Ans=Ans[::-1]
print(*Ans)
n=int(input())
P=[int(i) for i in input().split()]
import math
import bisect

n_max=2*10**5
nn=int(math.log2(n_max))+1
BIT=[0]*(2**nn+1)

def bitsum(BIT,i):
  s=0
  while i>0:
    s+=BIT[i]
    i-=i&(-i)
  return s
def bitadd(BIT,i,x):
  if i<=0:
    return True
  else:
    while i<=2**nn:
      BIT[i]+=x
      i+=i&(-i)
    return BIT
def bitlowerbound(BIT,s):
  if s<=0:
    return 0
  else:
    ret=0
    k=2**nn
    while k>0:
      if k+ret<=2**nn and BIT[k+ret]<s:
        s-=BIT[k+ret]
        ret+=k
      k//=2
    return ret+1
for i in range(n_max):
  bitadd(BIT,i+1,i+1)

Ans=[]
for i in reversed(range(n)):
  p=P[i]
  ans=bitlowerbound(BIT,p+1)
  Ans.append(ans)
  bitadd(BIT,ans,-ans)
Ans=Ans[::-1]
print(*Ans)