#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 200100;
const long long INF = 2000000100;
const long long MOD = 998244353;
long long N, S[MAXN], P[MAXN];
long long B[MAXN];
void update(int ind, long long v) {
  for (int i = ind; i <= N; i += (i & -i)) {
    B[i] += v;
  }
}
long long query(int ind) {
  if (ind == 0) return 0;
  long long s = 0;
  for (int i = ind; i > 0; i -= (i & -i)) {
    s += B[i];
  }
  return s;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> N;
  for (int i = 0; i < N; i++) cin >> S[i];
  for (int i = 1; i <= N; i++) update(i, i);
  for (int i = N - 1; i >= 0; i--) {
    long long a = 1, b = N, c;
    long long ans = -39;
    while (a <= b) {
      c = (a + b) / 2;
      if (query(c - 1) > S[i]) {
        b = c - 1;
      } else {
        ans = c;
        a = c + 1;
      }
    }
    P[i] = ans;
    update(ans, -ans);
  }
  for (int i = 0; i < N; i++) cout << P[i] << (i == N - 1 ? "\n" : " ");
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 MAXN = 200100;
const long long INF = 2000000100;
const long long MOD = 998244353;
long long N, S[MAXN], P[MAXN];
long long B[MAXN];
void update(int ind, long long v) {
  for (int i = ind; i <= N; i += (i & -i)) {
    B[i] += v;
  }
}
long long query(int ind) {
  if (ind == 0) return 0;
  long long s = 0;
  for (int i = ind; i > 0; i -= (i & -i)) {
    s += B[i];
  }
  return s;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> N;
  for (int i = 0; i < N; i++) cin >> S[i];
  for (int i = 1; i <= N; i++) update(i, i);
  for (int i = N - 1; i >= 0; i--) {
    long long a = 1, b = N, c;
    long long ans = -39;
    while (a <= b) {
      c = (a + b) / 2;
      if (query(c - 1) > S[i]) {
        b = c - 1;
      } else {
        ans = c;
        a = c + 1;
      }
    }
    P[i] = ans;
    update(ans, -ans);
  }
  for (int i = 0; i < N; i++) cout << P[i] << (i == N - 1 ? "\n" : " ");
}

```