#include <bits/stdc++.h>
using namespace std;
const int MX = 200005;
vector<int> v;
long long n, a[MX], w[MX], tree[MX * 4 + 5];
long long con(int num, int nodeL, int nodeR) {
  if (nodeL == nodeR) return tree[num] = w[nodeL];
  long long mid = (nodeL + nodeR) / 2;
  return tree[num] =
             con(num * 2, nodeL, mid) + con(num * 2 + 1, mid + 1, nodeR);
}
long long sum(int L, int R, int num, int nodeL, int nodeR) {
  if (nodeR < L || nodeL > R) return 0;
  if (nodeL >= L && nodeR <= R) return tree[num];
  int mid = (nodeL + nodeR) / 2;
  return sum(L, R, num * 2, nodeL, mid) +
         sum(L, R, num * 2 + 1, mid + 1, nodeR);
}
long long upd(int idx, int num, int nodeL, int nodeR) {
  if (nodeL > idx || nodeR < idx) return 0;
  if (nodeL == nodeR) return tree[num] = 0;
  int mid = (nodeL + nodeR) / 2;
  upd(idx, num * 2, nodeL, mid);
  upd(idx, num * 2 + 1, mid + 1, nodeR);
  if (nodeL <= idx && nodeR >= idx) {
    tree[num] = tree[num * 2] + tree[num * 2 + 1];
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) {
    w[i] = i + 1;
  }
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  con(1, 0, n - 1);
  for (int i = n - 1; i >= 0; i--) {
    int s = 0, e = n - 1;
    while (s <= e) {
      int m = (s + e) / 2;
      long long u = sum(0, m, 1, 0, n - 1);
      if (u <= a[i]) {
        s = m + 1;
      } else {
        e = m - 1;
      }
    }
    v.push_back(s + 1);
    upd(s, 1, 0, n - 1);
  }
  for (int i = v.size() - 1; i >= 0; i--) {
    cout << v[i] << " ";
  }
}

### Prompt

Create a solution in cpp for 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;
const int MX = 200005;
vector<int> v;
long long n, a[MX], w[MX], tree[MX * 4 + 5];
long long con(int num, int nodeL, int nodeR) {
  if (nodeL == nodeR) return tree[num] = w[nodeL];
  long long mid = (nodeL + nodeR) / 2;
  return tree[num] =
             con(num * 2, nodeL, mid) + con(num * 2 + 1, mid + 1, nodeR);
}
long long sum(int L, int R, int num, int nodeL, int nodeR) {
  if (nodeR < L || nodeL > R) return 0;
  if (nodeL >= L && nodeR <= R) return tree[num];
  int mid = (nodeL + nodeR) / 2;
  return sum(L, R, num * 2, nodeL, mid) +
         sum(L, R, num * 2 + 1, mid + 1, nodeR);
}
long long upd(int idx, int num, int nodeL, int nodeR) {
  if (nodeL > idx || nodeR < idx) return 0;
  if (nodeL == nodeR) return tree[num] = 0;
  int mid = (nodeL + nodeR) / 2;
  upd(idx, num * 2, nodeL, mid);
  upd(idx, num * 2 + 1, mid + 1, nodeR);
  if (nodeL <= idx && nodeR >= idx) {
    tree[num] = tree[num * 2] + tree[num * 2 + 1];
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) {
    w[i] = i + 1;
  }
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  con(1, 0, n - 1);
  for (int i = n - 1; i >= 0; i--) {
    int s = 0, e = n - 1;
    while (s <= e) {
      int m = (s + e) / 2;
      long long u = sum(0, m, 1, 0, n - 1);
      if (u <= a[i]) {
        s = m + 1;
      } else {
        e = m - 1;
      }
    }
    v.push_back(s + 1);
    upd(s, 1, 0, n - 1);
  }
  for (int i = v.size() - 1; i >= 0; i--) {
    cout << v[i] << " ";
  }
}

```