#include <bits/stdc++.h>
using namespace std;
long long n, ar[200005], br[200005], up[800555], dr[100];
pair<long long, long long> tr[800555];
void build(long long, long long, long long),
    update(long long, long long, long long, long long),
    upd(long long, long long, long long, long long, long long, long long);
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> ar[i];
  build(0, 0, n - 1);
  long long pos;
  for (int i = 1; i < n + 1; i++) {
    pos = tr[0].second;
    br[pos] = i;
    if (pos + 1 < n) upd(0, 0, n - 1, pos + 1, n - 1, i);
    update(0, 0, n - 1, pos);
  }
  for (int i = 0; i < n; i++) cout << br[i] << " ";
}
void build(long long idx, long long l, long long r) {
  if (l == r) {
    tr[idx].first = ar[l];
    tr[idx].second = l;
    return;
  }
  long long mid = (l + r) / 2;
  build(2 * idx + 1, l, mid);
  build(2 * idx + 2, mid + 1, r);
  if (tr[2 * idx + 1].first < tr[2 * idx + 2].first)
    tr[idx] = tr[2 * idx + 1];
  else
    tr[idx] = tr[2 * idx + 2];
}
void update(long long idx, long long l, long long r, long long pos) {
  if (up[idx]) {
    tr[idx].first -= up[idx];
    if (l != r) {
      up[2 * idx + 1] += up[idx];
      up[2 * idx + 2] += up[idx];
    }
    up[idx] = 0;
  }
  if (pos < l || pos > r) return;
  if (l == r) {
    tr[idx].first = 12345678123456ll;
    return;
  }
  long long mid = (l + r) / 2;
  update(2 * idx + 1, l, mid, pos);
  update(2 * idx + 2, mid + 1, r, pos);
  if (tr[2 * idx + 1].first < tr[2 * idx + 2].first)
    tr[idx] = tr[2 * idx + 1];
  else
    tr[idx] = tr[2 * idx + 2];
}
void upd(long long idx, long long l, long long r, long long low, long long high,
         long long val) {
  if (up[idx]) {
    tr[idx].first -= up[idx];
    if (l != r) {
      up[2 * idx + 1] += up[idx];
      up[2 * idx + 2] += up[idx];
    }
    up[idx] = 0;
  }
  if (high < l || low > r) {
    return;
  }
  if (low <= l && r <= high) {
    tr[idx].first -= val;
    if (l != r) {
      up[2 * idx + 1] += val;
      up[2 * idx + 2] += val;
    }
    return;
  }
  long long mid = (l + r) / 2;
  upd(2 * idx + 1, l, mid, low, high, val);
  upd(2 * idx + 2, mid + 1, r, low, high, val);
  if (tr[2 * idx + 1].first < tr[2 * idx + 2].first)
    tr[idx] = tr[2 * idx + 1];
  else
    tr[idx] = tr[2 * idx + 2];
}

### Prompt

Generate a CPP solution to the following problem:

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;
long long n, ar[200005], br[200005], up[800555], dr[100];
pair<long long, long long> tr[800555];
void build(long long, long long, long long),
    update(long long, long long, long long, long long),
    upd(long long, long long, long long, long long, long long, long long);
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> ar[i];
  build(0, 0, n - 1);
  long long pos;
  for (int i = 1; i < n + 1; i++) {
    pos = tr[0].second;
    br[pos] = i;
    if (pos + 1 < n) upd(0, 0, n - 1, pos + 1, n - 1, i);
    update(0, 0, n - 1, pos);
  }
  for (int i = 0; i < n; i++) cout << br[i] << " ";
}
void build(long long idx, long long l, long long r) {
  if (l == r) {
    tr[idx].first = ar[l];
    tr[idx].second = l;
    return;
  }
  long long mid = (l + r) / 2;
  build(2 * idx + 1, l, mid);
  build(2 * idx + 2, mid + 1, r);
  if (tr[2 * idx + 1].first < tr[2 * idx + 2].first)
    tr[idx] = tr[2 * idx + 1];
  else
    tr[idx] = tr[2 * idx + 2];
}
void update(long long idx, long long l, long long r, long long pos) {
  if (up[idx]) {
    tr[idx].first -= up[idx];
    if (l != r) {
      up[2 * idx + 1] += up[idx];
      up[2 * idx + 2] += up[idx];
    }
    up[idx] = 0;
  }
  if (pos < l || pos > r) return;
  if (l == r) {
    tr[idx].first = 12345678123456ll;
    return;
  }
  long long mid = (l + r) / 2;
  update(2 * idx + 1, l, mid, pos);
  update(2 * idx + 2, mid + 1, r, pos);
  if (tr[2 * idx + 1].first < tr[2 * idx + 2].first)
    tr[idx] = tr[2 * idx + 1];
  else
    tr[idx] = tr[2 * idx + 2];
}
void upd(long long idx, long long l, long long r, long long low, long long high,
         long long val) {
  if (up[idx]) {
    tr[idx].first -= up[idx];
    if (l != r) {
      up[2 * idx + 1] += up[idx];
      up[2 * idx + 2] += up[idx];
    }
    up[idx] = 0;
  }
  if (high < l || low > r) {
    return;
  }
  if (low <= l && r <= high) {
    tr[idx].first -= val;
    if (l != r) {
      up[2 * idx + 1] += val;
      up[2 * idx + 2] += val;
    }
    return;
  }
  long long mid = (l + r) / 2;
  upd(2 * idx + 1, l, mid, low, high, val);
  upd(2 * idx + 2, mid + 1, r, low, high, val);
  if (tr[2 * idx + 1].first < tr[2 * idx + 2].first)
    tr[idx] = tr[2 * idx + 1];
  else
    tr[idx] = tr[2 * idx + 2];
}

```