#include <bits/stdc++.h>
using namespace std;
long long int fen[200005];
void upd(long long int x, long long int a) {
  for (long long int i = x; i < 200005; i += (i & (-i))) {
    fen[i] += a;
  }
}
long long int ret(long long int x) {
  long long int o = 0;
  for (long long int i = x; i > 0; i -= (i & (-i))) {
    o += fen[i];
  }
  return o;
}
int main() {
  long long int n;
  cin >> n;
  long long int s[n];
  for (long long int i = 0; i < n; i++) {
    cin >> s[i];
    upd(i + 1, i + 1);
  }
  long long int out[n];
  for (long long int i = n - 1; i >= 0; i--) {
    long long int l = 0, h = n - 1;
    while (h - l > 1) {
      long long int mi = (h + l) / 2;
      if (ret(mi) > s[i]) {
        h = mi;
      } else {
        l = mi;
      }
    }
    if (ret(h) == s[i]) {
      out[i] = h + 1;
      upd(h + 1, -h - 1);
    } else {
      out[i] = l + 1;
      upd(l + 1, -l - 1);
    }
    cout << "\n";
  }
  for (long long int i = 0; i < n; i++) {
    cout << out[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;
long long int fen[200005];
void upd(long long int x, long long int a) {
  for (long long int i = x; i < 200005; i += (i & (-i))) {
    fen[i] += a;
  }
}
long long int ret(long long int x) {
  long long int o = 0;
  for (long long int i = x; i > 0; i -= (i & (-i))) {
    o += fen[i];
  }
  return o;
}
int main() {
  long long int n;
  cin >> n;
  long long int s[n];
  for (long long int i = 0; i < n; i++) {
    cin >> s[i];
    upd(i + 1, i + 1);
  }
  long long int out[n];
  for (long long int i = n - 1; i >= 0; i--) {
    long long int l = 0, h = n - 1;
    while (h - l > 1) {
      long long int mi = (h + l) / 2;
      if (ret(mi) > s[i]) {
        h = mi;
      } else {
        l = mi;
      }
    }
    if (ret(h) == s[i]) {
      out[i] = h + 1;
      upd(h + 1, -h - 1);
    } else {
      out[i] = l + 1;
      upd(l + 1, -l - 1);
    }
    cout << "\n";
  }
  for (long long int i = 0; i < n; i++) {
    cout << out[i] << " ";
  }
  return 0;
}

```