#include <bits/stdc++.h>
using namespace std;
int n, lo[(long long)(2e5 + 5) * 4], hi[(long long)(2e5 + 5) * 4],
    ans[(long long)(2e5 + 5)];
long long base[(long long)(2e5 + 5)], delta[(long long)(2e5 + 5) * 4];
pair<long long, int> mn[(long long)(2e5 + 5) * 4];
void update(int i);
void ini(int i, int a, int b) {
  lo[i] = a;
  hi[i] = b;
  if (a == b) {
    mn[i].first = base[a];
    mn[i].second = a;
    return;
  }
  int mid = (a + b) / 2;
  ini(2 * i, a, mid);
  ini(2 * i + 1, mid + 1, b);
  update(i);
}
void update(int i) {
  if (mn[i * 2].first + delta[i * 2] == mn[i * 2 + 1].first + delta[i * 2 + 1])
    mn[i] = mn[i * 2 + 1], mn[i].first += delta[i * 2 + 1];
  else if (mn[i * 2].first + delta[i * 2] <
           mn[i * 2 + 1].first + delta[i * 2 + 1])
    mn[i] = mn[i * 2], mn[i].first += delta[i * 2];
  else
    mn[i] = mn[i * 2 + 1], mn[i].first += delta[i * 2 + 1];
}
void prop(int i) {
  delta[i * 2] += delta[i];
  delta[i * 2 + 1] += delta[i];
  delta[i] = 0;
}
void inc(int i, int a, int b, long long val) {
  if (lo[i] >= a && hi[i] <= b) {
    delta[i] += val;
    return;
  }
  if (lo[i] > b || hi[i] < a) return;
  prop(i);
  inc(i * 2, a, b, val);
  inc(i * 2 + 1, a, b, val);
  update(i);
}
void change(int i, int loc) {
  if (lo[i] == loc && hi[i] == loc) {
    mn[i].first = 1e18, mn[i].second = -1;
    delta[i] = 1e18;
    return;
  }
  if (lo[i] > loc || hi[i] < loc) return;
  prop(i);
  change(i * 2, loc);
  change(i * 2 + 1, loc);
  update(i);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> base[i];
  ini(1, 1, n);
  for (int i = 1; i <= n; i++) {
    pair<long long, int> loc = mn[1];
    change(1, loc.second);
    inc(1, loc.second + 1, n, -1 * i);
    ans[loc.second] = i;
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i];
    if (i != n) cout << " ";
  }
  return 0;
}

### Prompt

Generate 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;
int n, lo[(long long)(2e5 + 5) * 4], hi[(long long)(2e5 + 5) * 4],
    ans[(long long)(2e5 + 5)];
long long base[(long long)(2e5 + 5)], delta[(long long)(2e5 + 5) * 4];
pair<long long, int> mn[(long long)(2e5 + 5) * 4];
void update(int i);
void ini(int i, int a, int b) {
  lo[i] = a;
  hi[i] = b;
  if (a == b) {
    mn[i].first = base[a];
    mn[i].second = a;
    return;
  }
  int mid = (a + b) / 2;
  ini(2 * i, a, mid);
  ini(2 * i + 1, mid + 1, b);
  update(i);
}
void update(int i) {
  if (mn[i * 2].first + delta[i * 2] == mn[i * 2 + 1].first + delta[i * 2 + 1])
    mn[i] = mn[i * 2 + 1], mn[i].first += delta[i * 2 + 1];
  else if (mn[i * 2].first + delta[i * 2] <
           mn[i * 2 + 1].first + delta[i * 2 + 1])
    mn[i] = mn[i * 2], mn[i].first += delta[i * 2];
  else
    mn[i] = mn[i * 2 + 1], mn[i].first += delta[i * 2 + 1];
}
void prop(int i) {
  delta[i * 2] += delta[i];
  delta[i * 2 + 1] += delta[i];
  delta[i] = 0;
}
void inc(int i, int a, int b, long long val) {
  if (lo[i] >= a && hi[i] <= b) {
    delta[i] += val;
    return;
  }
  if (lo[i] > b || hi[i] < a) return;
  prop(i);
  inc(i * 2, a, b, val);
  inc(i * 2 + 1, a, b, val);
  update(i);
}
void change(int i, int loc) {
  if (lo[i] == loc && hi[i] == loc) {
    mn[i].first = 1e18, mn[i].second = -1;
    delta[i] = 1e18;
    return;
  }
  if (lo[i] > loc || hi[i] < loc) return;
  prop(i);
  change(i * 2, loc);
  change(i * 2 + 1, loc);
  update(i);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> base[i];
  ini(1, 1, n);
  for (int i = 1; i <= n; i++) {
    pair<long long, int> loc = mn[1];
    change(1, loc.second);
    inc(1, loc.second + 1, n, -1 * i);
    ans[loc.second] = i;
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i];
    if (i != n) cout << " ";
  }
  return 0;
}

```