#include <bits/stdc++.h>
using namespace std;
long long int ar[4 * 200005];
void update(int a, int b, int l, int h, int p) {
  if (a < l || a > h) return;
  if (l == h) {
    ar[p] += b;
    return;
  }
  int m = (l + h) / 2;
  update(a, b, l, m, 2 * p);
  update(a, b, m + 1, h, 2 * p + 1);
  ar[p] += b;
  return;
}
int query(int l, int h, int p, long long int s) {
  if (l == h) return l;
  int m = (l + h) / 2;
  if (ar[2 * p] > s)
    return query(l, m, 2 * p, s);
  else
    return query(m + 1, h, 2 * p + 1, s - ar[2 * p]);
}
int compute(long long int s) {
  int x;
  s *= 2;
  x = sqrt(s);
  return x + 1;
}
int main() {
  int i, n;
  scanf("%d", &n);
  vector<long long int> sum(n);
  for (i = 0; i < n; i++) cin >> sum[i];
  vector<int> ans(n);
  ans[n - 1] = compute(sum[n - 1]);
  for (i = 1; i < n + 1; i++) update(i, i, 0, n, 1);
  update(ans[n - 1], -ans[n - 1], 0, n, 1);
  for (i = n - 2; i >= 0; i--) {
    ans[i] = query(0, n, 1, sum[i]);
    update(ans[i], -ans[i], 0, n, 1);
  }
  for (i = 0; i < n; i++) printf("%d ", ans[i]);
  printf("\n");
}

### Prompt

Please formulate 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 int ar[4 * 200005];
void update(int a, int b, int l, int h, int p) {
  if (a < l || a > h) return;
  if (l == h) {
    ar[p] += b;
    return;
  }
  int m = (l + h) / 2;
  update(a, b, l, m, 2 * p);
  update(a, b, m + 1, h, 2 * p + 1);
  ar[p] += b;
  return;
}
int query(int l, int h, int p, long long int s) {
  if (l == h) return l;
  int m = (l + h) / 2;
  if (ar[2 * p] > s)
    return query(l, m, 2 * p, s);
  else
    return query(m + 1, h, 2 * p + 1, s - ar[2 * p]);
}
int compute(long long int s) {
  int x;
  s *= 2;
  x = sqrt(s);
  return x + 1;
}
int main() {
  int i, n;
  scanf("%d", &n);
  vector<long long int> sum(n);
  for (i = 0; i < n; i++) cin >> sum[i];
  vector<int> ans(n);
  ans[n - 1] = compute(sum[n - 1]);
  for (i = 1; i < n + 1; i++) update(i, i, 0, n, 1);
  update(ans[n - 1], -ans[n - 1], 0, n, 1);
  for (i = n - 2; i >= 0; i--) {
    ans[i] = query(0, n, 1, sum[i]);
    update(ans[i], -ans[i], 0, n, 1);
  }
  for (i = 0; i < n; i++) printf("%d ", ans[i]);
  printf("\n");
}

```