#include <bits/stdc++.h>
using namespace std;
const long long MM = 10000000000000001;
int n;
long long a[200005];
int rs[200005];
long long lazy[4 * 200005];
pair<long long, int> it[4 * 200005];
pair<long long, int> minp(pair<long long, int> A, pair<long long, int> B) {
  if (A.first < B.first) return A;
  if (A.first > B.first) return B;
  if (A.second > B.second) return A;
  return B;
}
void build(int x, int l, int r) {
  if (l == r) {
    it[x] = make_pair(a[l], l);
    return;
  }
  int mid = (l + r) / 2;
  build(2 * x, l, mid);
  build(2 * x + 1, mid + 1, r);
  it[x] = minp(it[2 * x], it[2 * x + 1]);
}
void lazy_upd(int x, int l, int r) {
  it[x].first += lazy[x];
  if (l != r) {
    lazy[2 * x] += lazy[x];
    lazy[2 * x + 1] += lazy[x];
  }
  lazy[x] = 0;
}
void update(int x, int l, int r, int u, int v, long long val) {
  lazy_upd(x, l, r);
  if ((r < u) || (v < l)) return;
  if ((u <= l) && (r <= v)) {
    lazy[x] += val;
    lazy_upd(x, l, r);
    return;
  }
  int mid = (l + r) / 2;
  update(2 * x, l, mid, u, v, val);
  update(2 * x + 1, mid + 1, r, u, v, val);
  it[x] = minp(it[2 * x], it[2 * x + 1]);
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    pair<long long, int> p = it[1];
    int id = p.second;
    rs[id] = i;
    update(1, 1, n, id, id, MM);
    update(1, 1, n, id + 1, n, -i);
  }
  for (int i = 1; i <= n; i++) cout << rs[i] << ' ';
  return 0;
}

### Prompt

Please formulate 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 MM = 10000000000000001;
int n;
long long a[200005];
int rs[200005];
long long lazy[4 * 200005];
pair<long long, int> it[4 * 200005];
pair<long long, int> minp(pair<long long, int> A, pair<long long, int> B) {
  if (A.first < B.first) return A;
  if (A.first > B.first) return B;
  if (A.second > B.second) return A;
  return B;
}
void build(int x, int l, int r) {
  if (l == r) {
    it[x] = make_pair(a[l], l);
    return;
  }
  int mid = (l + r) / 2;
  build(2 * x, l, mid);
  build(2 * x + 1, mid + 1, r);
  it[x] = minp(it[2 * x], it[2 * x + 1]);
}
void lazy_upd(int x, int l, int r) {
  it[x].first += lazy[x];
  if (l != r) {
    lazy[2 * x] += lazy[x];
    lazy[2 * x + 1] += lazy[x];
  }
  lazy[x] = 0;
}
void update(int x, int l, int r, int u, int v, long long val) {
  lazy_upd(x, l, r);
  if ((r < u) || (v < l)) return;
  if ((u <= l) && (r <= v)) {
    lazy[x] += val;
    lazy_upd(x, l, r);
    return;
  }
  int mid = (l + r) / 2;
  update(2 * x, l, mid, u, v, val);
  update(2 * x + 1, mid + 1, r, u, v, val);
  it[x] = minp(it[2 * x], it[2 * x + 1]);
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    pair<long long, int> p = it[1];
    int id = p.second;
    rs[id] = i;
    update(1, 1, n, id, id, MM);
    update(1, 1, n, id + 1, n, -i);
  }
  for (int i = 1; i <= n; i++) cout << rs[i] << ' ';
  return 0;
}

```