#include <bits/stdc++.h>
const int MAXN = 2e5 + 10;
const long long inf = 1e17 + 10;
using namespace std;
struct Seg {
  long long v[MAXN * 4], lazy[MAXN * 4];
  int m(int l, int r) { return (l + r) / 2; }
  void updPoint(int pos, int l, int r, int x, long long val) {
    if (l == r) {
      v[pos] = val;
      return;
    }
    if (x <= m(l, r))
      updPoint(pos * 2, l, m(l, r), x, val);
    else
      updPoint(pos * 2 + 1, m(l, r) + 1, r, x, val);
    v[pos] = min(v[pos * 2], v[pos * 2 + 1]);
  }
  void refresh(int pos, int l, int r) {
    long long k = lazy[pos];
    lazy[pos] = 0;
    v[pos] -= k;
    if (l == r) return;
    for (int i = 0; i < 2; i++) lazy[pos * 2 + i] += k;
  }
  void updInterval(int pos, int l, int r, int beg, int en, long long val) {
    refresh(pos, l, r);
    if (l > en || r < beg) return;
    if (l >= beg && r <= en) {
      lazy[pos] += val;
      refresh(pos, l, r);
      return;
    }
    updInterval(pos * 2, l, m(l, r), beg, en, val);
    updInterval(pos * 2 + 1, m(l, r) + 1, r, beg, en, val);
    v[pos] = min(v[pos * 2], v[pos * 2 + 1]);
  }
  int buscab(int pos, int l, int r) {
    refresh(pos, l, r);
    if (l == r) return l;
    refresh(pos * 2, l, m(l, r));
    refresh(pos * 2 + 1, m(l, r) + 1, r);
    if (v[pos * 2] > 0) return buscab(pos * 2 + 1, m(l, r) + 1, r);
    return buscab(pos * 2, l, m(l, r));
  }
};
int n;
long long vet[MAXN];
int seq[MAXN];
Seg seg;
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n + 1; i++) scanf("%lld", &vet[i]);
  reverse(vet + 1, vet + 1 + n);
  for (int i = 1; i < n + 1; i++) seg.updPoint(1, 1, n, i, vet[i]);
  for (int i = 1; i < n + 1; i++) {
    int t = seg.buscab(1, 1, n);
    seg.updInterval(1, 1, n, 1, t - 1, i);
    seg.updInterval(1, 1, n, t, t, -inf);
    seq[n - t + 1] = i;
  }
  for (int i = 1; i < n + 1; i++) printf("%d ", seq[i]);
  printf("\n");
}

### Prompt

In CPP, your task is to solve 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>
const int MAXN = 2e5 + 10;
const long long inf = 1e17 + 10;
using namespace std;
struct Seg {
  long long v[MAXN * 4], lazy[MAXN * 4];
  int m(int l, int r) { return (l + r) / 2; }
  void updPoint(int pos, int l, int r, int x, long long val) {
    if (l == r) {
      v[pos] = val;
      return;
    }
    if (x <= m(l, r))
      updPoint(pos * 2, l, m(l, r), x, val);
    else
      updPoint(pos * 2 + 1, m(l, r) + 1, r, x, val);
    v[pos] = min(v[pos * 2], v[pos * 2 + 1]);
  }
  void refresh(int pos, int l, int r) {
    long long k = lazy[pos];
    lazy[pos] = 0;
    v[pos] -= k;
    if (l == r) return;
    for (int i = 0; i < 2; i++) lazy[pos * 2 + i] += k;
  }
  void updInterval(int pos, int l, int r, int beg, int en, long long val) {
    refresh(pos, l, r);
    if (l > en || r < beg) return;
    if (l >= beg && r <= en) {
      lazy[pos] += val;
      refresh(pos, l, r);
      return;
    }
    updInterval(pos * 2, l, m(l, r), beg, en, val);
    updInterval(pos * 2 + 1, m(l, r) + 1, r, beg, en, val);
    v[pos] = min(v[pos * 2], v[pos * 2 + 1]);
  }
  int buscab(int pos, int l, int r) {
    refresh(pos, l, r);
    if (l == r) return l;
    refresh(pos * 2, l, m(l, r));
    refresh(pos * 2 + 1, m(l, r) + 1, r);
    if (v[pos * 2] > 0) return buscab(pos * 2 + 1, m(l, r) + 1, r);
    return buscab(pos * 2, l, m(l, r));
  }
};
int n;
long long vet[MAXN];
int seq[MAXN];
Seg seg;
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n + 1; i++) scanf("%lld", &vet[i]);
  reverse(vet + 1, vet + 1 + n);
  for (int i = 1; i < n + 1; i++) seg.updPoint(1, 1, n, i, vet[i]);
  for (int i = 1; i < n + 1; i++) {
    int t = seg.buscab(1, 1, n);
    seg.updInterval(1, 1, n, 1, t - 1, i);
    seg.updInterval(1, 1, n, t, t, -inf);
    seq[n - t + 1] = i;
  }
  for (int i = 1; i < n + 1; i++) printf("%d ", seq[i]);
  printf("\n");
}

```