#include <bits/stdc++.h>
using namespace std;
int n;
long long s[200001];
int used[200001];
long long stree[800001];
int result[200001];
long long update(int node, int l, int r, int index, long long diff) {
  if (r < index || index < l) return stree[node];
  if (r == l) {
    stree[node] += diff;
    return stree[node];
  };
  return stree[node] = update(node * 2, l, (l + r) / 2, index, diff) +
                       update(node * 2 + 1, (l + r) / 2 + 1, r, index, diff);
}
long long query(int node, int l, int r, int left, int right) {
  if (r < left || l > right) return 0;
  if (l >= left && r <= right) return stree[node];
  return query(node * 2, l, (l + r) / 2, left, right) +
         query(node * 2 + 1, (l + r) / 2 + 1, r, left, right);
}
int bsearch(long long c) {
  int mi = 1;
  int ma = n;
  if (c == 0 && used[1] == 0) return 0;
  while (mi + 1 < ma) {
    int mid = (mi + ma) / 2;
    long long res = query(1, 1, n, 1, mid);
    if (res <= c)
      mi = mid;
    else
      ma = mid;
  }
  return mi;
}
int main(void) {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
  }
  for (int i = 1; i <= n; i++) {
    update(1, 1, n, i, i);
  }
  for (int i = n; i >= 1; i--) {
    result[i] = bsearch(s[i]) + 1;
    update(1, 1, n, result[i], -result[i]);
    used[result[i]] = 1;
  }
  for (int i = 1; i <= n; i++) {
    cout << result[i] << " ";
  }
}

### Prompt

Your challenge is to write 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;
int n;
long long s[200001];
int used[200001];
long long stree[800001];
int result[200001];
long long update(int node, int l, int r, int index, long long diff) {
  if (r < index || index < l) return stree[node];
  if (r == l) {
    stree[node] += diff;
    return stree[node];
  };
  return stree[node] = update(node * 2, l, (l + r) / 2, index, diff) +
                       update(node * 2 + 1, (l + r) / 2 + 1, r, index, diff);
}
long long query(int node, int l, int r, int left, int right) {
  if (r < left || l > right) return 0;
  if (l >= left && r <= right) return stree[node];
  return query(node * 2, l, (l + r) / 2, left, right) +
         query(node * 2 + 1, (l + r) / 2 + 1, r, left, right);
}
int bsearch(long long c) {
  int mi = 1;
  int ma = n;
  if (c == 0 && used[1] == 0) return 0;
  while (mi + 1 < ma) {
    int mid = (mi + ma) / 2;
    long long res = query(1, 1, n, 1, mid);
    if (res <= c)
      mi = mid;
    else
      ma = mid;
  }
  return mi;
}
int main(void) {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
  }
  for (int i = 1; i <= n; i++) {
    update(1, 1, n, i, i);
  }
  for (int i = n; i >= 1; i--) {
    result[i] = bsearch(s[i]) + 1;
    update(1, 1, n, result[i], -result[i]);
    used[result[i]] = 1;
  }
  for (int i = 1; i <= n; i++) {
    cout << result[i] << " ";
  }
}

```