#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10, eps = 1e-10;
int n;
long long a[N];
long long tree[N];
int ans[N];
long long sum(int x) {
  long long res = 0;
  while (x) {
    res += tree[x];
    x -= (x & (-x));
  }
  return res;
}
void add(int x, long long sum) {
  while (x <= n) {
    tree[x] += sum;
    x += (x & (-x));
  }
  return;
}
int main() {
  std::ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = (1); i < (n + 1); ++i) cin >> a[i], add(i + 1, i);
  for (int i = (n); i >= (1); --i) {
    long long l = 1, r = n;
    while (l <= r) {
      if (sum(((l + r) >> 1)) > a[i])
        r = ((l + r) >> 1) - 1;
      else
        l = ((l + r) >> 1) + 1;
    }
    ans[i] = r;
    add(r + 1, -r);
  }
  for (int i = (1); i < (n + 1); ++i) cout << ans[i] << " ";
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

An array of integers p_{1},p_{2}, …,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 ≤ n ≤ 2 ⋅ 10^{5}) — the size of the permutation.

The second line contains n integers s_{1}, s_{2}, …, s_{n} (0 ≤ s_{i} ≤ (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}, …, 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10, eps = 1e-10;
int n;
long long a[N];
long long tree[N];
int ans[N];
long long sum(int x) {
  long long res = 0;
  while (x) {
    res += tree[x];
    x -= (x & (-x));
  }
  return res;
}
void add(int x, long long sum) {
  while (x <= n) {
    tree[x] += sum;
    x += (x & (-x));
  }
  return;
}
int main() {
  std::ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = (1); i < (n + 1); ++i) cin >> a[i], add(i + 1, i);
  for (int i = (n); i >= (1); --i) {
    long long l = 1, r = n;
    while (l <= r) {
      if (sum(((l + r) >> 1)) > a[i])
        r = ((l + r) >> 1) - 1;
      else
        l = ((l + r) >> 1) + 1;
    }
    ans[i] = r;
    add(r + 1, -r);
  }
  for (int i = (1); i < (n + 1); ++i) cout << ans[i] << " ";
  return 0;
}

```