#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int n;
long long sum[N], ans[N], cur;
long long seg[1 << 20];
set<int> nw;
void update(int i, int v, int ni = 0, int ns = 0, int ne = n) {
  if (ns > i || ne < i || ns > ne) return;
  if (ns == ne && ns == i) {
    seg[ni] += v;
    return;
  }
  if (ns >= ne) return;
  int l = ni * 2 + 1, r = l + 1, mid = (ns + ne) / 2;
  update(i, v, l, ns, mid);
  update(i, v, r, mid + 1, ne);
  seg[ni] = seg[l] + seg[r];
}
long long query(int qs, int qe, int ni = 0, int ns = 0, int ne = n) {
  if (ns > qe || ne < qs || ns > ne) return 0;
  if (ns >= qs && ne <= qe) return seg[ni];
  int l = ni * 2 + 1, r = l + 1, mid = (ns + ne) / 2;
  return query(qs, qe, l, ns, mid) + query(qs, qe, r, mid + 1, ne);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%lld", sum + i);
  for (int i = 1; i <= n; ++i) nw.insert(i);
  for (int i = n - 1; i >= 0; --i) {
    int st = 0, en = n - 1;
    while (st < en) {
      int mid = (st + en + 1) / 2;
      long long s = 1ll * mid * (mid + 1) / 2 - query(0, mid);
      if (s > sum[i])
        en = mid - 1;
      else
        st = mid;
    }
    int x = *nw.lower_bound(st + 1);
    nw.erase(x);
    ans[i] = x;
    update(x, x);
  }
  for (int i = 0; i < n; ++i) printf("%lld ", ans[i]);
  return 0;
}

### Prompt

Your task is to create 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;
const int N = 2e5 + 5;
int n;
long long sum[N], ans[N], cur;
long long seg[1 << 20];
set<int> nw;
void update(int i, int v, int ni = 0, int ns = 0, int ne = n) {
  if (ns > i || ne < i || ns > ne) return;
  if (ns == ne && ns == i) {
    seg[ni] += v;
    return;
  }
  if (ns >= ne) return;
  int l = ni * 2 + 1, r = l + 1, mid = (ns + ne) / 2;
  update(i, v, l, ns, mid);
  update(i, v, r, mid + 1, ne);
  seg[ni] = seg[l] + seg[r];
}
long long query(int qs, int qe, int ni = 0, int ns = 0, int ne = n) {
  if (ns > qe || ne < qs || ns > ne) return 0;
  if (ns >= qs && ne <= qe) return seg[ni];
  int l = ni * 2 + 1, r = l + 1, mid = (ns + ne) / 2;
  return query(qs, qe, l, ns, mid) + query(qs, qe, r, mid + 1, ne);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%lld", sum + i);
  for (int i = 1; i <= n; ++i) nw.insert(i);
  for (int i = n - 1; i >= 0; --i) {
    int st = 0, en = n - 1;
    while (st < en) {
      int mid = (st + en + 1) / 2;
      long long s = 1ll * mid * (mid + 1) / 2 - query(0, mid);
      if (s > sum[i])
        en = mid - 1;
      else
        st = mid;
    }
    int x = *nw.lower_bound(st + 1);
    nw.erase(x);
    ans[i] = x;
    update(x, x);
  }
  for (int i = 0; i < n; ++i) printf("%lld ", ans[i]);
  return 0;
}

```