#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll P[200001];
ll sum[200001];
ll fen[200001];
int ans[200001];
void build() {
  for (int i = 1; i <= 200000; i++) {
    sum[i] = sum[i - 1] + ll(i - 1);
    fen[i] = sum[i] - sum[i - (i & -i)];
  }
}
ll query(int l, int r) {
  if (l != 1LL) return query(1, r) - query(1, l - 1);
  ll ans = 0LL;
  while (r > 0) {
    ans += fen[r];
    r -= r & -r;
  }
  return ans;
}
void update(int i) {
  ll taking = ll(i);
  i++;
  while (i <= 200000) {
    fen[i] -= taking;
    i += (i & -i);
  }
}
int find(ll s) {
  ll cur = 0LL;
  int ind = 0;
  for (int b = 17; b >= 0; b--) {
    if (ind + (1 << b) <= 200000 && cur + fen[ind + (1 << b)] <= s) {
      cur += fen[ind + (1 << b)];
      ind += (1 << b);
    }
  }
  return ind;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> P[i];
  build();
  for (int i = n; i >= 1; i--) {
    ans[i] = find(P[i]);
    update(ans[i]);
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
}

### Prompt

In CPP, your task is to solve 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;
using ll = long long;
ll P[200001];
ll sum[200001];
ll fen[200001];
int ans[200001];
void build() {
  for (int i = 1; i <= 200000; i++) {
    sum[i] = sum[i - 1] + ll(i - 1);
    fen[i] = sum[i] - sum[i - (i & -i)];
  }
}
ll query(int l, int r) {
  if (l != 1LL) return query(1, r) - query(1, l - 1);
  ll ans = 0LL;
  while (r > 0) {
    ans += fen[r];
    r -= r & -r;
  }
  return ans;
}
void update(int i) {
  ll taking = ll(i);
  i++;
  while (i <= 200000) {
    fen[i] -= taking;
    i += (i & -i);
  }
}
int find(ll s) {
  ll cur = 0LL;
  int ind = 0;
  for (int b = 17; b >= 0; b--) {
    if (ind + (1 << b) <= 200000 && cur + fen[ind + (1 << b)] <= s) {
      cur += fen[ind + (1 << b)];
      ind += (1 << b);
    }
  }
  return ind;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> P[i];
  build();
  for (int i = n; i >= 1; i--) {
    ans[i] = find(P[i]);
    update(ans[i]);
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
}

```