#include <bits/stdc++.h>
std::vector<long long> a;
const long long INF = 1000000000000000000LL;
int lm = -1;
std::vector<long long> ts;
long long fill(int l, int r, int i) {
  if (l == r) {
    ts[i] = a[l];
    return a[l];
  }
  int m = (l + r) / 2;
  ts[i] = fill(l, m, i * 2) + fill(m + 1, r, i * 2 + 1);
  return ts[i];
}
void init(int n) {
  ts.assign(4 * n, 0);
  fill(0, n - 1, 1);
}
long long getSum(int l, int r, int cl, int cr, int i) {
  if (l == cl && r == cr) return ts[i];
  int cm = (cl + cr) / 2;
  if (r <= cm) return getSum(l, r, cl, cm, i * 2);
  if (cm < l) return getSum(l, r, cm + 1, cr, i * 2 + 1);
  return getSum(l, cm, cl, cm, i * 2) +
         getSum(cm + 1, r, cm + 1, cr, i * 2 + 1);
}
void set(int cl, int cr, int i, int ai, int val) {
  if (cl == ai && cr == ai) {
    ts[i] = a[ai] = val;
    return;
  }
  int cm = (cl + cr) / 2;
  if (ai <= cm)
    set(cl, cm, i * 2, ai, val);
  else
    set(cm + 1, cr, i * 2 + 1, ai, val);
  ts[i] = ts[i * 2] + ts[i * 2 + 1];
}
int main() {
  int n;
  scanf("%d", &n);
  std::vector<long long> s(n);
  for (long long& x : s) scanf("%I64d", &x);
  a.assign(1 + n, 0);
  for (int i = 0; i < (int)a.size(); ++i) a[i] = i;
  init(a.size());
  std::vector<int> ans(n);
  for (int i = n - 1; i >= 0; --i) {
    int l = 0, r = n + 1;
    while (r > l + 1) {
      int m = (l + r) / 2;
      long long sum = getSum(0, m, 0, n, 1);
      if (sum > s[i])
        r = m;
      else
        l = m;
    }
    ans[i] = l + 1;
    set(0, n, 1, l + 1, 0);
  }
  for (int x : ans) printf("%d ", x);
  return 0;
}

### 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>
std::vector<long long> a;
const long long INF = 1000000000000000000LL;
int lm = -1;
std::vector<long long> ts;
long long fill(int l, int r, int i) {
  if (l == r) {
    ts[i] = a[l];
    return a[l];
  }
  int m = (l + r) / 2;
  ts[i] = fill(l, m, i * 2) + fill(m + 1, r, i * 2 + 1);
  return ts[i];
}
void init(int n) {
  ts.assign(4 * n, 0);
  fill(0, n - 1, 1);
}
long long getSum(int l, int r, int cl, int cr, int i) {
  if (l == cl && r == cr) return ts[i];
  int cm = (cl + cr) / 2;
  if (r <= cm) return getSum(l, r, cl, cm, i * 2);
  if (cm < l) return getSum(l, r, cm + 1, cr, i * 2 + 1);
  return getSum(l, cm, cl, cm, i * 2) +
         getSum(cm + 1, r, cm + 1, cr, i * 2 + 1);
}
void set(int cl, int cr, int i, int ai, int val) {
  if (cl == ai && cr == ai) {
    ts[i] = a[ai] = val;
    return;
  }
  int cm = (cl + cr) / 2;
  if (ai <= cm)
    set(cl, cm, i * 2, ai, val);
  else
    set(cm + 1, cr, i * 2 + 1, ai, val);
  ts[i] = ts[i * 2] + ts[i * 2 + 1];
}
int main() {
  int n;
  scanf("%d", &n);
  std::vector<long long> s(n);
  for (long long& x : s) scanf("%I64d", &x);
  a.assign(1 + n, 0);
  for (int i = 0; i < (int)a.size(); ++i) a[i] = i;
  init(a.size());
  std::vector<int> ans(n);
  for (int i = n - 1; i >= 0; --i) {
    int l = 0, r = n + 1;
    while (r > l + 1) {
      int m = (l + r) / 2;
      long long sum = getSum(0, m, 0, n, 1);
      if (sum > s[i])
        r = m;
      else
        l = m;
    }
    ans[i] = l + 1;
    set(0, n, 1, l + 1, 0);
  }
  for (int x : ans) printf("%d ", x);
  return 0;
}

```