#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
const int N = 2e5 + 20;
int n, s[N], a[N], fen[N];
inline void add(int p, int val) {
  for (p++; p < N; p += p & -p) fen[p] += val;
}
inline int get(int p) {
  int sum = 0;
  for (; p; p -= p & -p) sum += fen[p];
  return sum;
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> s[i];
  for (int i = 0; i < n; i++) add(i + 1, i + 1);
  int p = n - 1;
  while (p >= 0) {
    int low = 0, high = n + 1;
    while (high - low > 1) {
      int mid = (low + high) >> 1;
      if (get(mid) <= s[p])
        low = mid;
      else
        high = mid;
    }
    a[p] = low;
    add(low, -low);
    p--;
  }
  for (int i = 0; i < n; i++) cout << a[i] << ' ';
  cout << '\n';
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
const int N = 2e5 + 20;
int n, s[N], a[N], fen[N];
inline void add(int p, int val) {
  for (p++; p < N; p += p & -p) fen[p] += val;
}
inline int get(int p) {
  int sum = 0;
  for (; p; p -= p & -p) sum += fen[p];
  return sum;
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> s[i];
  for (int i = 0; i < n; i++) add(i + 1, i + 1);
  int p = n - 1;
  while (p >= 0) {
    int low = 0, high = n + 1;
    while (high - low > 1) {
      int mid = (low + high) >> 1;
      if (get(mid) <= s[p])
        low = mid;
      else
        high = mid;
    }
    a[p] = low;
    add(low, -low);
    p--;
  }
  for (int i = 0; i < n; i++) cout << a[i] << ' ';
  cout << '\n';
}

```