#include <bits/stdc++.h>
using namespace std;
template <class T>
bool mini(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool maxi(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
const int N = 2e5 + 5;
int n, ans[N];
long long bit[N], p[N];
void up(int i) {
  assert(i > 0);
  for (long long val = i; i <= n; i += i & -i) bit[i] += val;
}
long long get(int i) {
  long long res = 0;
  for (; i > 0; i -= i & -i) res += bit[i];
  return res;
}
inline void sol() {
  cin >> n;
  for (int i = 0, _ = (n); i < _; i++) cin >> p[i];
  for (int i = (int)(n)-1; i >= 0; --i) {
    long long l = 0, r = n - 1, mid;
    while (l < r) {
      mid = (l + r + 1) >> 1;
      if (mid * (mid + 1) / 2 > get(mid) + p[i])
        r = mid - 1;
      else
        l = mid;
    }
    up(ans[i] = l + 1);
  }
  for (int i = 0, _ = (n); i < _; i++) cout << ans[i] << ' ';
  cout << "\n";
}
signed main() {
  { ios_base::sync_with_stdio(false), cin.tie(NULL); };
  cout.precision(10);
  cout << fixed;
  sol();
}

### 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;
template <class T>
bool mini(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool maxi(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
const int N = 2e5 + 5;
int n, ans[N];
long long bit[N], p[N];
void up(int i) {
  assert(i > 0);
  for (long long val = i; i <= n; i += i & -i) bit[i] += val;
}
long long get(int i) {
  long long res = 0;
  for (; i > 0; i -= i & -i) res += bit[i];
  return res;
}
inline void sol() {
  cin >> n;
  for (int i = 0, _ = (n); i < _; i++) cin >> p[i];
  for (int i = (int)(n)-1; i >= 0; --i) {
    long long l = 0, r = n - 1, mid;
    while (l < r) {
      mid = (l + r + 1) >> 1;
      if (mid * (mid + 1) / 2 > get(mid) + p[i])
        r = mid - 1;
      else
        l = mid;
    }
    up(ans[i] = l + 1);
  }
  for (int i = 0, _ = (n); i < _; i++) cout << ans[i] << ' ';
  cout << "\n";
}
signed main() {
  { ios_base::sync_with_stdio(false), cin.tie(NULL); };
  cout.precision(10);
  cout << fixed;
  sol();
}

```