#include <bits/stdc++.h>
using namespace std;
long long a[3000000], tree[4 * 3000000], ans[4 * 3000000];
int n;
void update(int indx, int left, int right, int pos, long long val) {
  if (left > pos || right < pos) return;
  if (left == right) {
    tree[indx] += val;
    return;
  }
  int mid = (left + right) >> 1;
  update(indx * 2, left, mid, pos, val);
  update(indx * 2 + 1, mid + 1, right, pos, val);
  tree[indx] = tree[indx * 2] + tree[indx * 2 + 1];
}
long long query(int indx, int l, int r, int left, int right) {
  if (left > right || l > right || r < left) return 0;
  if (l >= left && r <= right) return tree[indx];
  int mid = (l + r) >> 1;
  long long x = query(indx * 2, l, mid, left, right);
  long long y = query(indx * 2 + 1, mid + 1, r, left, right);
  return x + y;
}
int doit(long long val) {
  int low = 1, high = n, ans = 1;
  while (low <= high) {
    int mid = (low + high) >> 1;
    long long xx = query(1, 1, n, 1, mid - 1);
    if (xx <= val) {
      low = mid + 1;
      ans = mid;
    } else
      high = mid - 1;
  }
  return ans;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout.setf(ios::fixed);
  cout.precision(20);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) update(1, 1, n, i, i);
  for (int i = n; i > 0; i--) {
    ans[i] = doit(a[i]);
    update(1, 1, n, ans[i], -1ll * ans[i]);
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  return 0;
}

### Prompt

Create a solution in cpp for 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 a[3000000], tree[4 * 3000000], ans[4 * 3000000];
int n;
void update(int indx, int left, int right, int pos, long long val) {
  if (left > pos || right < pos) return;
  if (left == right) {
    tree[indx] += val;
    return;
  }
  int mid = (left + right) >> 1;
  update(indx * 2, left, mid, pos, val);
  update(indx * 2 + 1, mid + 1, right, pos, val);
  tree[indx] = tree[indx * 2] + tree[indx * 2 + 1];
}
long long query(int indx, int l, int r, int left, int right) {
  if (left > right || l > right || r < left) return 0;
  if (l >= left && r <= right) return tree[indx];
  int mid = (l + r) >> 1;
  long long x = query(indx * 2, l, mid, left, right);
  long long y = query(indx * 2 + 1, mid + 1, r, left, right);
  return x + y;
}
int doit(long long val) {
  int low = 1, high = n, ans = 1;
  while (low <= high) {
    int mid = (low + high) >> 1;
    long long xx = query(1, 1, n, 1, mid - 1);
    if (xx <= val) {
      low = mid + 1;
      ans = mid;
    } else
      high = mid - 1;
  }
  return ans;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout.setf(ios::fixed);
  cout.precision(20);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) update(1, 1, n, i, i);
  for (int i = n; i > 0; i--) {
    ans[i] = doit(a[i]);
    update(1, 1, n, ans[i], -1ll * ans[i]);
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  return 0;
}

```