#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
long long n = 1, n2, s[200010], ans[200010], dat[800010];
void INIT(long long k, long long lb, long long ub) {
  if (lb == ub) return;
  INIT(k * 2 + 1, lb, (lb + ub) / 2);
  INIT(k * 2 + 2, (lb + ub) / 2 + 1, ub);
  dat[k] = dat[k * 2 + 1] + dat[k * 2 + 2];
}
long long query(long long k, long long lb, long long ub, long long tlb,
                long long tub) {
  if (ub < tlb || tub < lb) return 0;
  if (tlb <= lb && ub <= tub) return dat[k];
  return query(k * 2 + 1, lb, (lb + ub) / 2, tlb, tub) +
         query(k * 2 + 2, (lb + ub) / 2 + 1, ub, tlb, tub);
}
void upd(long long val) {
  long long k = val - 1 + n - 1;
  dat[k] -= val;
  while (k > 0) {
    dat[(k - 1) / 2] -= val;
    k = (k - 1) / 2;
  }
}
bool check(long long mid, long long i) {
  return query(0, 0, n - 1, 0, mid - 2) <= s[i];
}
int main() {
  cin >> n2;
  while (n < n2) n *= 2;
  for (int i = 0; i < n2; i++) scanf("%I64d", &s[i]);
  for (int i = 0; i < n; i++) dat[i + n - 1] = i + 1;
  INIT(0, 0, n - 1);
  for (int i = n2 - 1; i >= 0; i--) {
    int lb = 1, ub = n2, mid;
    while (lb < ub) {
      mid = (lb + ub + 1) / 2;
      if (check(mid, i))
        lb = mid;
      else
        ub = mid - 1;
    }
    ans[i] = lb;
    upd(lb);
  }
  check(2, 2);
  for (int i = 0; i < n2; i++) cout << ans[i] << ' ';
  cout << endl;
  return 0;
}

### Prompt

Your task is to create 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;
const long long MOD = 1e9 + 7;
long long n = 1, n2, s[200010], ans[200010], dat[800010];
void INIT(long long k, long long lb, long long ub) {
  if (lb == ub) return;
  INIT(k * 2 + 1, lb, (lb + ub) / 2);
  INIT(k * 2 + 2, (lb + ub) / 2 + 1, ub);
  dat[k] = dat[k * 2 + 1] + dat[k * 2 + 2];
}
long long query(long long k, long long lb, long long ub, long long tlb,
                long long tub) {
  if (ub < tlb || tub < lb) return 0;
  if (tlb <= lb && ub <= tub) return dat[k];
  return query(k * 2 + 1, lb, (lb + ub) / 2, tlb, tub) +
         query(k * 2 + 2, (lb + ub) / 2 + 1, ub, tlb, tub);
}
void upd(long long val) {
  long long k = val - 1 + n - 1;
  dat[k] -= val;
  while (k > 0) {
    dat[(k - 1) / 2] -= val;
    k = (k - 1) / 2;
  }
}
bool check(long long mid, long long i) {
  return query(0, 0, n - 1, 0, mid - 2) <= s[i];
}
int main() {
  cin >> n2;
  while (n < n2) n *= 2;
  for (int i = 0; i < n2; i++) scanf("%I64d", &s[i]);
  for (int i = 0; i < n; i++) dat[i + n - 1] = i + 1;
  INIT(0, 0, n - 1);
  for (int i = n2 - 1; i >= 0; i--) {
    int lb = 1, ub = n2, mid;
    while (lb < ub) {
      mid = (lb + ub + 1) / 2;
      if (check(mid, i))
        lb = mid;
      else
        ub = mid - 1;
    }
    ans[i] = lb;
    upd(lb);
  }
  check(2, 2);
  for (int i = 0; i < n2; i++) cout << ans[i] << ' ';
  cout << endl;
  return 0;
}

```