#include <bits/stdc++.h>
using namespace std;
using LL = long long;
using PII = pair<int, int>;
template <class T>
struct Fenwick {
  vector<T> v;
  Fenwick(size_t n) : v(n + 1) {}
  void add(size_t i, T x) {
    for (++i; i < v.size(); i += i & -i) v[i] += x;
  }
  T sum(size_t i) {
    for (v[0] = T(); i; i -= i & -i) v[0] += v.at(i);
    return v[0];
  }
  T sum(size_t l, size_t r) { return sum(r) - sum(l); }
};
const int N = 2e5 + 5;
int n, a[N];
LL s[N];
int main() {
  cin >> n;
  Fenwick<LL> fw(N);
  for (int i = (1); i <= (n); ++i) scanf("%lld", &s[i]), fw.add(i, i);
  for (int i = n; i >= 1; i--) {
    int l = 0, r = n;
    while (l + 1 < r) {
      int m = (l + r) >> 1;
      LL sum = fw.sum(m + 1);
      if (sum <= s[i])
        l = m;
      else
        r = m;
    }
    a[i] = l + 1;
    fw.add(l + 1, -(l + 1));
  }
  for (int i = 1; i <= n; i++) printf("%d%c", a[i], " \n"[i == n]);
}

### 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;
using LL = long long;
using PII = pair<int, int>;
template <class T>
struct Fenwick {
  vector<T> v;
  Fenwick(size_t n) : v(n + 1) {}
  void add(size_t i, T x) {
    for (++i; i < v.size(); i += i & -i) v[i] += x;
  }
  T sum(size_t i) {
    for (v[0] = T(); i; i -= i & -i) v[0] += v.at(i);
    return v[0];
  }
  T sum(size_t l, size_t r) { return sum(r) - sum(l); }
};
const int N = 2e5 + 5;
int n, a[N];
LL s[N];
int main() {
  cin >> n;
  Fenwick<LL> fw(N);
  for (int i = (1); i <= (n); ++i) scanf("%lld", &s[i]), fw.add(i, i);
  for (int i = n; i >= 1; i--) {
    int l = 0, r = n;
    while (l + 1 < r) {
      int m = (l + r) >> 1;
      LL sum = fw.sum(m + 1);
      if (sum <= s[i])
        l = m;
      else
        r = m;
    }
    a[i] = l + 1;
    fw.add(l + 1, -(l + 1));
  }
  for (int i = 1; i <= n; i++) printf("%d%c", a[i], " \n"[i == n]);
}

```