#include <bits/stdc++.h>
using namespace std;
struct BIT {
  int n, N_MAX;
  vector<long long> v;
  BIT(int n) {
    this->n = n + 100;
    N_MAX = n;
    v.assign(n + 110, 0);
  }
  void upd(int p, int x) {
    while (p <= n) v[p] += x, p += p & -p;
  }
  long long que(int p) {
    long long ans = 0;
    while (p) ans += v[p], p -= p & -p;
    return ans;
  }
  long long quep(int p) { return que(p) - que(p - 1); }
  long long bit_search(long long s) {
    long long sum = 0, pos = 0;
    for (int i = 21; i >= 0; i--)
      if (pos + (1 << i) <= N_MAX && sum + v[pos + (1 << i)] <= s) {
        pos += (1 << i);
        sum += v[pos];
      }
    return pos + 1;
  }
};
int main() {
  int n;
  scanf("%d", &n);
  BIT bit(n);
  vector<long long> v(n + 1), ans(n + 1);
  for (int i = 1; i <= n; i++) scanf("%lld", &v[i]), bit.upd(i, i);
  for (int i = n; i; i--) {
    int p = bit.bit_search(v[i]);
    ans[i] = p;
    bit.upd(p, -p);
  }
  for (int i = 1; i <= n; i++) printf("%lld ", ans[i]);
  printf("\n");
  return 0;
}

### 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;
struct BIT {
  int n, N_MAX;
  vector<long long> v;
  BIT(int n) {
    this->n = n + 100;
    N_MAX = n;
    v.assign(n + 110, 0);
  }
  void upd(int p, int x) {
    while (p <= n) v[p] += x, p += p & -p;
  }
  long long que(int p) {
    long long ans = 0;
    while (p) ans += v[p], p -= p & -p;
    return ans;
  }
  long long quep(int p) { return que(p) - que(p - 1); }
  long long bit_search(long long s) {
    long long sum = 0, pos = 0;
    for (int i = 21; i >= 0; i--)
      if (pos + (1 << i) <= N_MAX && sum + v[pos + (1 << i)] <= s) {
        pos += (1 << i);
        sum += v[pos];
      }
    return pos + 1;
  }
};
int main() {
  int n;
  scanf("%d", &n);
  BIT bit(n);
  vector<long long> v(n + 1), ans(n + 1);
  for (int i = 1; i <= n; i++) scanf("%lld", &v[i]), bit.upd(i, i);
  for (int i = n; i; i--) {
    int p = bit.bit_search(v[i]);
    ans[i] = p;
    bit.upd(p, -p);
  }
  for (int i = 1; i <= n; i++) printf("%lld ", ans[i]);
  printf("\n");
  return 0;
}

```