#include <bits/stdc++.h>
using namespace std;
using pii = pair<long long, int>;
const int maxn = 2e5 + 5;
int n;
long long a[maxn], s[maxn], t[maxn << 2];
void build(int l, int r, int rt) {
  if (l == r) {
    t[rt] = l;
    return;
  }
  int m = l + r >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  t[rt] = t[rt << 1] + t[rt << 1 | 1];
}
void update(int pos, int l, int r, int rt) {
  if (l > pos || r < pos) return;
  if (l == r) {
    t[rt] = 0;
    return;
  }
  int m = l + r >> 1;
  update(pos, l, m, rt << 1);
  update(pos, m + 1, r, rt << 1 | 1);
  t[rt] = t[rt << 1] + t[rt << 1 | 1];
}
int query(long long q, int l, int r, int rt) {
  if (l == r) return l;
  int m = l + r >> 1;
  if (t[rt << 1] > q)
    return query(q, l, m, rt << 1);
  else
    return query(q - t[rt << 1], m + 1, r, rt << 1 | 1);
}
void pri(int l, int r, int rt) {
  if (l == r) {
    printf("%lld  ", t[rt]);
    return;
  }
  int m = l + r >> 1;
  pri(l, m, rt << 1);
  pri(m + 1, r, rt << 1 | 1);
}
int main() {
  scanf("%d", &n);
  build(1, n, 1);
  for (int i = 1; i <= n; i++) scanf("%lld", s + i);
  for (int i = n; i > 0; i--) {
    a[i] = query(s[i], 1, n, 1);
    update(a[i], 1, n, 1);
  }
  for (int i = 1; i <= n; i++) printf("%lld ", a[i]);
}

### 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;
using pii = pair<long long, int>;
const int maxn = 2e5 + 5;
int n;
long long a[maxn], s[maxn], t[maxn << 2];
void build(int l, int r, int rt) {
  if (l == r) {
    t[rt] = l;
    return;
  }
  int m = l + r >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  t[rt] = t[rt << 1] + t[rt << 1 | 1];
}
void update(int pos, int l, int r, int rt) {
  if (l > pos || r < pos) return;
  if (l == r) {
    t[rt] = 0;
    return;
  }
  int m = l + r >> 1;
  update(pos, l, m, rt << 1);
  update(pos, m + 1, r, rt << 1 | 1);
  t[rt] = t[rt << 1] + t[rt << 1 | 1];
}
int query(long long q, int l, int r, int rt) {
  if (l == r) return l;
  int m = l + r >> 1;
  if (t[rt << 1] > q)
    return query(q, l, m, rt << 1);
  else
    return query(q - t[rt << 1], m + 1, r, rt << 1 | 1);
}
void pri(int l, int r, int rt) {
  if (l == r) {
    printf("%lld  ", t[rt]);
    return;
  }
  int m = l + r >> 1;
  pri(l, m, rt << 1);
  pri(m + 1, r, rt << 1 | 1);
}
int main() {
  scanf("%d", &n);
  build(1, n, 1);
  for (int i = 1; i <= n; i++) scanf("%lld", s + i);
  for (int i = n; i > 0; i--) {
    a[i] = query(s[i], 1, n, 1);
    update(a[i], 1, n, 1);
  }
  for (int i = 1; i <= n; i++) printf("%lld ", a[i]);
}

```