#include <bits/stdc++.h>
using namespace std;
int T;
long long n;
long long a[333333];
long long twosum[333333];
long long tree[(1LL << 18) * 3];
long long ans[333333];
set<long long> s;
void update(long long x, long long y) {
  x += (1LL << 18);
  while (x > 0LL) {
    tree[x] += y;
    x /= 2LL;
  }
}
long long sum(int s, int e, int ns = 0, int ne = (1LL << 18) - 1, int idx = 1) {
  if (s > ne || ns > e) return 0LL;
  if (s <= ns && ne <= e) return tree[idx];
  int mid = (ns + ne) / 2;
  return sum(s, e, ns, mid, idx * 2) + sum(s, e, mid + 1, ne, idx * 2 + 1);
}
int main() {
  scanf("%lld", &n);
  for (long long i = 1LL; i <= n; i++) {
    scanf("%lld", &a[i]);
    twosum[i] = twosum[i - 1] + i;
    s.insert(i);
    update(i, i);
  }
  for (long long i = n; i > 0; i--) {
    long long st = 1LL;
    long long ed = n;
    long long val;
    while (st <= ed) {
      long long mid = (st + ed) / 2LL;
      if (sum(1, (int)mid - 1) <= a[i]) {
        st = mid + 1LL;
        val = mid;
      } else
        ed = mid - 1LL;
    }
    val = *s.lower_bound(val);
    update(val, -val);
    s.erase(val);
    ans[i] = val;
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
int T;
long long n;
long long a[333333];
long long twosum[333333];
long long tree[(1LL << 18) * 3];
long long ans[333333];
set<long long> s;
void update(long long x, long long y) {
  x += (1LL << 18);
  while (x > 0LL) {
    tree[x] += y;
    x /= 2LL;
  }
}
long long sum(int s, int e, int ns = 0, int ne = (1LL << 18) - 1, int idx = 1) {
  if (s > ne || ns > e) return 0LL;
  if (s <= ns && ne <= e) return tree[idx];
  int mid = (ns + ne) / 2;
  return sum(s, e, ns, mid, idx * 2) + sum(s, e, mid + 1, ne, idx * 2 + 1);
}
int main() {
  scanf("%lld", &n);
  for (long long i = 1LL; i <= n; i++) {
    scanf("%lld", &a[i]);
    twosum[i] = twosum[i - 1] + i;
    s.insert(i);
    update(i, i);
  }
  for (long long i = n; i > 0; i--) {
    long long st = 1LL;
    long long ed = n;
    long long val;
    while (st <= ed) {
      long long mid = (st + ed) / 2LL;
      if (sum(1, (int)mid - 1) <= a[i]) {
        st = mid + 1LL;
        val = mid;
      } else
        ed = mid - 1LL;
    }
    val = *s.lower_bound(val);
    update(val, -val);
    s.erase(val);
    ans[i] = val;
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}

```