#include <bits/stdc++.h>
using namespace std;
long long n;
class fenwicktree {
 public:
  long long n;
  long long b[200005];
  fenwicktree(long long N) {
    n = N;
    for (long long i = 0; i <= n; i++) b[i] = 0;
  }
  long long sum(long long idx) {
    long long ret = 0;
    for (idx; idx > 0; idx -= (idx & -idx)) ret += b[idx];
    return ret;
  }
  void add(long long idx, long long delta) {
    for (idx; idx <= n; idx += (idx & -idx)) b[idx] += delta;
  }
};
long long bin(long long l, long long r, long long x, fenwicktree &f) {
  while (l < r) {
    long long mid = (l + r + 1) / 2;
    if (f.sum(mid) <= x)
      l = mid;
    else
      r = mid - 1;
  }
  return l + 1;
}
void solve() {
  cin >> n;
  fenwicktree f(n + 5);
  for (long long i = 1; i <= n; i++) f.add(i, i);
  long long a[200005];
  long long res[200005];
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = n - 1; i >= 0; i--) {
    long long x = bin(0, n - 1, a[i], f);
    res[i] = x;
    f.add(x, -x);
  }
  for (long long i = 0; i < n; i++) cout << res[i] << " ";
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  t = 1;
  while (t--) solve();
}

### Prompt

Please create a solution in CPP 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 n;
class fenwicktree {
 public:
  long long n;
  long long b[200005];
  fenwicktree(long long N) {
    n = N;
    for (long long i = 0; i <= n; i++) b[i] = 0;
  }
  long long sum(long long idx) {
    long long ret = 0;
    for (idx; idx > 0; idx -= (idx & -idx)) ret += b[idx];
    return ret;
  }
  void add(long long idx, long long delta) {
    for (idx; idx <= n; idx += (idx & -idx)) b[idx] += delta;
  }
};
long long bin(long long l, long long r, long long x, fenwicktree &f) {
  while (l < r) {
    long long mid = (l + r + 1) / 2;
    if (f.sum(mid) <= x)
      l = mid;
    else
      r = mid - 1;
  }
  return l + 1;
}
void solve() {
  cin >> n;
  fenwicktree f(n + 5);
  for (long long i = 1; i <= n; i++) f.add(i, i);
  long long a[200005];
  long long res[200005];
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = n - 1; i >= 0; i--) {
    long long x = bin(0, n - 1, a[i], f);
    res[i] = x;
    f.add(x, -x);
  }
  for (long long i = 0; i < n; i++) cout << res[i] << " ";
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  t = 1;
  while (t--) solve();
}

```