#include <bits/stdc++.h>
#pragma GCC optimize "O3"
using namespace std;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int N = 2e5 + 5;
const int M = 1 << 18;
pair<long long, int> tree[M * 2 + 5];
long long lazy[M * 2 + 5];
int n, ans[N];
void push(int v) {
  tree[v * 2].first += lazy[v];
  tree[v * 2 + 1].first += lazy[v];
  lazy[v * 2] += lazy[v];
  lazy[v * 2 + 1] += lazy[v];
  lazy[v] = 0;
}
void add(int l, int r, long long val, int v = 1, int le = 1, int re = M - 1) {
  if (l > r || l > re || r < le) return;
  if (le >= l && re <= r) {
    tree[v].first += val;
    lazy[v] += val;
    return;
  }
  int mid = (le + re) / 2;
  push(v);
  add(l, r, val, v * 2, le, mid);
  add(l, r, val, v * 2 + 1, mid + 1, re);
  tree[v] = min(tree[v * 2], tree[v * 2 + 1]);
}
void solve() {
  int pom;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> tree[i + M - 1].first;
    tree[i + M - 1].second = {-i};
  }
  for (int i = M - 1; i >= 1; i--) tree[i] = min(tree[i * 2], tree[i * 2 + 1]);
  for (int i = 1; i <= n; i++) {
    pom = -tree[1].second;
    ans[pom] = i;
    add(pom, pom, INF);
    add(pom + 1, n, -i);
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

### Prompt

Develop a solution in cpp 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>
#pragma GCC optimize "O3"
using namespace std;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int N = 2e5 + 5;
const int M = 1 << 18;
pair<long long, int> tree[M * 2 + 5];
long long lazy[M * 2 + 5];
int n, ans[N];
void push(int v) {
  tree[v * 2].first += lazy[v];
  tree[v * 2 + 1].first += lazy[v];
  lazy[v * 2] += lazy[v];
  lazy[v * 2 + 1] += lazy[v];
  lazy[v] = 0;
}
void add(int l, int r, long long val, int v = 1, int le = 1, int re = M - 1) {
  if (l > r || l > re || r < le) return;
  if (le >= l && re <= r) {
    tree[v].first += val;
    lazy[v] += val;
    return;
  }
  int mid = (le + re) / 2;
  push(v);
  add(l, r, val, v * 2, le, mid);
  add(l, r, val, v * 2 + 1, mid + 1, re);
  tree[v] = min(tree[v * 2], tree[v * 2 + 1]);
}
void solve() {
  int pom;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> tree[i + M - 1].first;
    tree[i + M - 1].second = {-i};
  }
  for (int i = M - 1; i >= 1; i--) tree[i] = min(tree[i * 2], tree[i * 2 + 1]);
  for (int i = 1; i <= n; i++) {
    pom = -tree[1].second;
    ans[pom] = i;
    add(pom, pom, INF);
    add(pom + 1, n, -i);
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

```