#include <bits/stdc++.h>
using namespace std;
int n, a[200005];
long long s[200005];
pair<long long, int> sg[800005];
long long lz[800005];
void Build(int nod, int l, int r) {
  sg[nod] = {0, r};
  if (l == r) return;
  int mid = (l + r) / 2;
  Build(2 * nod, l, mid);
  Build(2 * nod + 1, mid + 1, r);
}
void Sift(int nod, int l, int r) {
  sg[nod].first += lz[nod];
  if (l != r) {
    lz[2 * nod] += lz[nod];
    lz[2 * nod + 1] += lz[nod];
  }
  lz[nod] = 0;
}
void Upd(int nod, int l, int r, int lt, int rt, long long v) {
  Sift(nod, l, r);
  if (l > rt || r < lt) return;
  if (l >= lt && r <= rt) {
    lz[nod] += v;
    Sift(nod, l, r);
    return;
  }
  int mid = (l + r) / 2;
  Upd(2 * nod, l, mid, lt, rt, v);
  Upd(2 * nod + 1, mid + 1, r, lt, rt, v);
  if (sg[2 * nod + 1].first > sg[2 * nod].first)
    sg[nod] = sg[2 * nod];
  else
    sg[nod] = sg[2 * nod + 1];
}
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  cin >> n;
  Build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    Upd(1, 1, n, i, i, s[i]);
  }
  for (int i = 1; i <= n; i++) {
    int p = sg[1].second;
    a[p] = i;
    Upd(1, 1, n, p, n, -i);
    Upd(1, 1, n, p, p, 1000000000000000005);
  }
  for (int i = 1; i <= n; i++) cout << a[i] << ' ';
}

### Prompt

Please provide a cpp coded solution 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;
int n, a[200005];
long long s[200005];
pair<long long, int> sg[800005];
long long lz[800005];
void Build(int nod, int l, int r) {
  sg[nod] = {0, r};
  if (l == r) return;
  int mid = (l + r) / 2;
  Build(2 * nod, l, mid);
  Build(2 * nod + 1, mid + 1, r);
}
void Sift(int nod, int l, int r) {
  sg[nod].first += lz[nod];
  if (l != r) {
    lz[2 * nod] += lz[nod];
    lz[2 * nod + 1] += lz[nod];
  }
  lz[nod] = 0;
}
void Upd(int nod, int l, int r, int lt, int rt, long long v) {
  Sift(nod, l, r);
  if (l > rt || r < lt) return;
  if (l >= lt && r <= rt) {
    lz[nod] += v;
    Sift(nod, l, r);
    return;
  }
  int mid = (l + r) / 2;
  Upd(2 * nod, l, mid, lt, rt, v);
  Upd(2 * nod + 1, mid + 1, r, lt, rt, v);
  if (sg[2 * nod + 1].first > sg[2 * nod].first)
    sg[nod] = sg[2 * nod];
  else
    sg[nod] = sg[2 * nod + 1];
}
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  cin >> n;
  Build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    Upd(1, 1, n, i, i, s[i]);
  }
  for (int i = 1; i <= n; i++) {
    int p = sg[1].second;
    a[p] = i;
    Upd(1, 1, n, p, n, -i);
    Upd(1, 1, n, p, p, 1000000000000000005);
  }
  for (int i = 1; i <= n; i++) cout << a[i] << ' ';
}

```