#include <bits/stdc++.h>
using namespace std;
long long t[200025 * 4], lz[200025 * 4], n, a[200025];
void build(int l, int r, int i) {
  t[i] = 1ll * l * (l - 1) / 2;
  if (l == r) return;
  build(l, (l + r) / 2, i * 2 + 1);
  build((l + r) / 2 + 1, r, i * 2 + 2);
  t[i] = max(t[i * 2 + 1], t[i * 2 + 2]);
}
void down(int i) {
  lz[i * 2 + 1] += lz[i];
  lz[i * 2 + 2] += lz[i];
  t[i * 2 + 1] += lz[i];
  t[i * 2 + 2] += lz[i];
  lz[i] = 0;
}
void updata(int L, int R, long long v, int l, int r, int i) {
  if (L <= l && R >= r) {
    t[i] += v;
    lz[i] += v;
    return;
  }
  if (lz[i]) down(i);
  if (L <= (l + r) / 2) updata(L, R, v, l, (l + r) / 2, i * 2 + 1);
  if (R > (l + r) / 2) updata(L, R, v, (l + r) / 2 + 1, r, i * 2 + 2);
  t[i] = max(t[i * 2 + 1], t[i * 2 + 2]);
}
int query(long long v, int l, int r, int i) {
  if (l == r) return l;
  if (lz[i]) down(i);
  if (t[i * 2 + 1] >= v) return query(v, l, (l + r) / 2, i * 2 + 1);
  return query(v, (l + r) / 2 + 1, r, i * 2 + 2);
}
int main() {
  cin >> n;
  for (int i = (1); i <= (n); i++) scanf("%lld", &a[i]);
  build(1, n, 0);
  for (int i = (n); i >= (1); i--) {
    a[i] = query(a[i], 1, n, 0);
    if (a[i] < n) updata(a[i] + 1, n, -a[i], 1, n, 0);
    updata(a[i], a[i], -1e12, 1, n, 0);
  }
  for (int i = (1); i <= (n); i++) printf("%lld%c", a[i], i == n ? '\n' : ' ');
  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;
long long t[200025 * 4], lz[200025 * 4], n, a[200025];
void build(int l, int r, int i) {
  t[i] = 1ll * l * (l - 1) / 2;
  if (l == r) return;
  build(l, (l + r) / 2, i * 2 + 1);
  build((l + r) / 2 + 1, r, i * 2 + 2);
  t[i] = max(t[i * 2 + 1], t[i * 2 + 2]);
}
void down(int i) {
  lz[i * 2 + 1] += lz[i];
  lz[i * 2 + 2] += lz[i];
  t[i * 2 + 1] += lz[i];
  t[i * 2 + 2] += lz[i];
  lz[i] = 0;
}
void updata(int L, int R, long long v, int l, int r, int i) {
  if (L <= l && R >= r) {
    t[i] += v;
    lz[i] += v;
    return;
  }
  if (lz[i]) down(i);
  if (L <= (l + r) / 2) updata(L, R, v, l, (l + r) / 2, i * 2 + 1);
  if (R > (l + r) / 2) updata(L, R, v, (l + r) / 2 + 1, r, i * 2 + 2);
  t[i] = max(t[i * 2 + 1], t[i * 2 + 2]);
}
int query(long long v, int l, int r, int i) {
  if (l == r) return l;
  if (lz[i]) down(i);
  if (t[i * 2 + 1] >= v) return query(v, l, (l + r) / 2, i * 2 + 1);
  return query(v, (l + r) / 2 + 1, r, i * 2 + 2);
}
int main() {
  cin >> n;
  for (int i = (1); i <= (n); i++) scanf("%lld", &a[i]);
  build(1, n, 0);
  for (int i = (n); i >= (1); i--) {
    a[i] = query(a[i], 1, n, 0);
    if (a[i] < n) updata(a[i] + 1, n, -a[i], 1, n, 0);
    updata(a[i], a[i], -1e12, 1, n, 0);
  }
  for (int i = (1); i <= (n); i++) printf("%lld%c", a[i], i == n ? '\n' : ' ');
  return 0;
}

```