#include <bits/stdc++.h>
using namespace std;
pair<long long, long long> t[4 * 200009];
long long a[200009], lazy[4 * 200009], ans[200009];
void build(long long v, long long tl, long long tr) {
  if (tl == tr) {
    t[v] = {a[tl], tl};
  } else {
    long long tm = (tl + tr) / 2;
    build(v * 2, tl, tm);
    build(v * 2 + 1, tm + 1, tr);
    if (t[v * 2].first < t[v * 2 + 1].first ||
        (t[v * 2].first == t[v * 2 + 1].first &&
         t[v * 2].second > t[v * 2 + 1].second))
      t[v] = t[v * 2];
    else
      t[v] = t[v * 2 + 1];
  }
}
void push(long long v) {
  t[v * 2].first += lazy[v];
  lazy[v * 2] += lazy[v];
  t[v * 2 + 1].first += lazy[v];
  lazy[v * 2 + 1] += lazy[v];
  lazy[v] = 0;
}
void update(long long v, long long tl, long long tr, long long l, long long r,
            long long addend) {
  if (l > r) return;
  if (l == tl && tr == r) {
    t[v].first += addend;
    lazy[v] += addend;
  } else {
    push(v);
    long long tm = (tl + tr) / 2;
    update(v * 2, tl, tm, l, min(r, tm), addend);
    update(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r, addend);
    if (t[v * 2].first < t[v * 2 + 1].first ||
        (t[v * 2].first == t[v * 2 + 1].first &&
         t[v * 2].second > t[v * 2 + 1].second))
      t[v] = t[v * 2];
    else
      t[v] = t[v * 2 + 1];
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  long long i = 1;
  while (i <= n) cin >> a[i++];
  build(1, 1, n);
  i = 1;
  long long x;
  while (i <= n) {
    x = t[1].second;
    ans[x] = i;
    update(1, 1, n, x + 1, n, -i);
    update(1, 1, n, x, x, 1e13);
    i++;
  }
  i = 1;
  while (i <= n) {
    cout << ans[i] << " ";
    i++;
  }
}

### 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;
pair<long long, long long> t[4 * 200009];
long long a[200009], lazy[4 * 200009], ans[200009];
void build(long long v, long long tl, long long tr) {
  if (tl == tr) {
    t[v] = {a[tl], tl};
  } else {
    long long tm = (tl + tr) / 2;
    build(v * 2, tl, tm);
    build(v * 2 + 1, tm + 1, tr);
    if (t[v * 2].first < t[v * 2 + 1].first ||
        (t[v * 2].first == t[v * 2 + 1].first &&
         t[v * 2].second > t[v * 2 + 1].second))
      t[v] = t[v * 2];
    else
      t[v] = t[v * 2 + 1];
  }
}
void push(long long v) {
  t[v * 2].first += lazy[v];
  lazy[v * 2] += lazy[v];
  t[v * 2 + 1].first += lazy[v];
  lazy[v * 2 + 1] += lazy[v];
  lazy[v] = 0;
}
void update(long long v, long long tl, long long tr, long long l, long long r,
            long long addend) {
  if (l > r) return;
  if (l == tl && tr == r) {
    t[v].first += addend;
    lazy[v] += addend;
  } else {
    push(v);
    long long tm = (tl + tr) / 2;
    update(v * 2, tl, tm, l, min(r, tm), addend);
    update(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r, addend);
    if (t[v * 2].first < t[v * 2 + 1].first ||
        (t[v * 2].first == t[v * 2 + 1].first &&
         t[v * 2].second > t[v * 2 + 1].second))
      t[v] = t[v * 2];
    else
      t[v] = t[v * 2 + 1];
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  long long i = 1;
  while (i <= n) cin >> a[i++];
  build(1, 1, n);
  i = 1;
  long long x;
  while (i <= n) {
    x = t[1].second;
    ans[x] = i;
    update(1, 1, n, x + 1, n, -i);
    update(1, 1, n, x, x, 1e13);
    i++;
  }
  i = 1;
  while (i <= n) {
    cout << ans[i] << " ";
    i++;
  }
}

```