#include <bits/stdc++.h>
using namespace std;
vector<long long> s;
vector<pair<long long, long long>> tree;
void build(long long l, long long r, long long v) {
  if (r - 1 == l) {
    tree[v] = {s[l], l};
    return;
  }
  long long m = (r + l) / 2;
  build(l, m, v * 2 + 1);
  build(m, r, v * 2 + 2);
  auto a = tree[v * 2 + 1];
  auto b = tree[v * 2 + 2];
  if (a.first < b.first) {
    tree[v] = a;
  } else {
    tree[v] = b;
  }
}
vector<long long> mod;
pair<long long, long long> get(long long l, long long r, long long u,
                               long long w, long long v) {
  if (r <= u || l >= w) {
    return {1e9, 0};
  }
  if (r <= w && l >= u) {
    return tree[v];
  }
  long long m = (r + l) / 2;
  auto a = get(l, m, u, w, v * 2 + 1);
  auto b = get(m, r, u, w, v * 2 + 2);
  if (a.first < b.first) {
    return {a.first + mod[v], a.second};
  } else {
    return {b.first + mod[v], b.second};
  }
}
void update(long long l, long long r, long long u, long long w, long long v,
            long long delt) {
  if (r <= u || l >= w) {
    return;
  }
  if (r <= w && l >= u) {
    mod[v] += delt;
    auto a = tree[v];
    tree[v] = {a.first + delt, a.second};
    return;
  }
  long long m = (r + l) / 2;
  update(l, m, u, w, v * 2 + 1, delt);
  update(m, r, u, w, v * 2 + 2, delt);
  auto a = tree[v * 2 + 1];
  auto b = tree[v * 2 + 2];
  if (a.first < b.first) {
    tree[v] = {a.first + mod[v], a.second};
  } else {
    tree[v] = {b.first + mod[v], b.second};
  }
}
signed main() {
  long long n;
  cin >> n;
  s.resize(n);
  for (long long i = 0; i < n; i++) {
    cin >> s[i];
  }
  tree.resize(4 * n);
  mod.resize(4 * n);
  build(0, n, 0);
  vector<long long> v(n);
  long long sum = 0;
  for (long long i = 1; i <= n; i++) {
    auto a = get(0, n, 0, n, 0);
    v[a.second] = i;
    update(0, n, a.second, a.second + 1, 0, 1e18 + 1);
    update(0, n, a.second + 1, n, 0, -i);
  }
  for (long long i = 0; i < n; i++) {
    cout << v[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;
vector<long long> s;
vector<pair<long long, long long>> tree;
void build(long long l, long long r, long long v) {
  if (r - 1 == l) {
    tree[v] = {s[l], l};
    return;
  }
  long long m = (r + l) / 2;
  build(l, m, v * 2 + 1);
  build(m, r, v * 2 + 2);
  auto a = tree[v * 2 + 1];
  auto b = tree[v * 2 + 2];
  if (a.first < b.first) {
    tree[v] = a;
  } else {
    tree[v] = b;
  }
}
vector<long long> mod;
pair<long long, long long> get(long long l, long long r, long long u,
                               long long w, long long v) {
  if (r <= u || l >= w) {
    return {1e9, 0};
  }
  if (r <= w && l >= u) {
    return tree[v];
  }
  long long m = (r + l) / 2;
  auto a = get(l, m, u, w, v * 2 + 1);
  auto b = get(m, r, u, w, v * 2 + 2);
  if (a.first < b.first) {
    return {a.first + mod[v], a.second};
  } else {
    return {b.first + mod[v], b.second};
  }
}
void update(long long l, long long r, long long u, long long w, long long v,
            long long delt) {
  if (r <= u || l >= w) {
    return;
  }
  if (r <= w && l >= u) {
    mod[v] += delt;
    auto a = tree[v];
    tree[v] = {a.first + delt, a.second};
    return;
  }
  long long m = (r + l) / 2;
  update(l, m, u, w, v * 2 + 1, delt);
  update(m, r, u, w, v * 2 + 2, delt);
  auto a = tree[v * 2 + 1];
  auto b = tree[v * 2 + 2];
  if (a.first < b.first) {
    tree[v] = {a.first + mod[v], a.second};
  } else {
    tree[v] = {b.first + mod[v], b.second};
  }
}
signed main() {
  long long n;
  cin >> n;
  s.resize(n);
  for (long long i = 0; i < n; i++) {
    cin >> s[i];
  }
  tree.resize(4 * n);
  mod.resize(4 * n);
  build(0, n, 0);
  vector<long long> v(n);
  long long sum = 0;
  for (long long i = 1; i <= n; i++) {
    auto a = get(0, n, 0, n, 0);
    v[a.second] = i;
    update(0, n, a.second, a.second + 1, 0, 1e18 + 1);
    update(0, n, a.second + 1, n, 0, -i);
  }
  for (long long i = 0; i < n; i++) {
    cout << v[i] << " ";
  }
  return 0;
}

```