#include <bits/stdc++.h>
using namespace std;
vector<unsigned long long> t, lazy;
void build(unsigned long long a[], int v, int tl, int tr) {
  if (tl == tr) {
    t[v] = a[tl];
  } else {
    int tm = (tl + tr) / 2;
    build(a, v * 2 + 1, tl, tm);
    build(a, v * 2 + 2, tm + 1, tr);
    t[v] = min(t[v * 2 + 1], t[v * 2 + 2]);
  }
}
void add(int v, int tl, int tr, int l, int r, unsigned long long addend) {
  if (l > r) return;
  if (l == tl && tr == r) {
    t[v] += addend;
    lazy[v] += addend;
  } else {
    t[v * 2 + 1] += lazy[v];
    lazy[v * 2 + 1] += lazy[v];
    t[v * 2 + 2] += lazy[v];
    lazy[v * 2 + 2] += lazy[v];
    lazy[v] = 0;
    int tm = (tl + tr) / 2;
    add(v * 2 + 1, tl, tm, l, min(r, tm), addend);
    add(v * 2 + 2, tm + 1, tr, max(l, tm + 1), r, addend);
    t[v] = min(t[v * 2 + 1], t[v * 2 + 2]);
  }
}
int last_zero(int v, int l, int r) {
  if (l == r) {
    return l;
  } else {
    int m = (r + l) / 2;
    t[v * 2 + 1] += lazy[v];
    lazy[v * 2 + 1] += lazy[v];
    t[v * 2 + 2] += lazy[v];
    lazy[v * 2 + 2] += lazy[v];
    lazy[v] = 0;
    t[v] = min(t[v * 2 + 1], t[v * 2 + 2]);
    if (t[2 * v + 2]) {
      return last_zero(2 * v + 1, l, m);
    }
    return last_zero(2 * v + 2, m + 1, r);
  }
}
int main() {
  const unsigned long long INF = 1e16 + 239;
  int n;
  cin >> n;
  unsigned long long s[n];
  int p[n];
  t.assign(n * 4, INF);
  lazy.assign(n * 4, 0);
  for (int i = 0; i < n; i++) {
    cin >> s[i];
  }
  build(s, 0, 0, n - 1);
  for (int i = 1; i <= n; i++) {
    int pos = last_zero(0, 0, n - 1);
    p[pos] = i;
    add(0, 0, n - 1, pos, pos, INF);
    add(0, 0, n - 1, pos + 1, n - 1, -i);
  }
  for (int i = 0; i < n; i++) {
    cout << p[i] << ' ';
  }
}

### 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>
using namespace std;
vector<unsigned long long> t, lazy;
void build(unsigned long long a[], int v, int tl, int tr) {
  if (tl == tr) {
    t[v] = a[tl];
  } else {
    int tm = (tl + tr) / 2;
    build(a, v * 2 + 1, tl, tm);
    build(a, v * 2 + 2, tm + 1, tr);
    t[v] = min(t[v * 2 + 1], t[v * 2 + 2]);
  }
}
void add(int v, int tl, int tr, int l, int r, unsigned long long addend) {
  if (l > r) return;
  if (l == tl && tr == r) {
    t[v] += addend;
    lazy[v] += addend;
  } else {
    t[v * 2 + 1] += lazy[v];
    lazy[v * 2 + 1] += lazy[v];
    t[v * 2 + 2] += lazy[v];
    lazy[v * 2 + 2] += lazy[v];
    lazy[v] = 0;
    int tm = (tl + tr) / 2;
    add(v * 2 + 1, tl, tm, l, min(r, tm), addend);
    add(v * 2 + 2, tm + 1, tr, max(l, tm + 1), r, addend);
    t[v] = min(t[v * 2 + 1], t[v * 2 + 2]);
  }
}
int last_zero(int v, int l, int r) {
  if (l == r) {
    return l;
  } else {
    int m = (r + l) / 2;
    t[v * 2 + 1] += lazy[v];
    lazy[v * 2 + 1] += lazy[v];
    t[v * 2 + 2] += lazy[v];
    lazy[v * 2 + 2] += lazy[v];
    lazy[v] = 0;
    t[v] = min(t[v * 2 + 1], t[v * 2 + 2]);
    if (t[2 * v + 2]) {
      return last_zero(2 * v + 1, l, m);
    }
    return last_zero(2 * v + 2, m + 1, r);
  }
}
int main() {
  const unsigned long long INF = 1e16 + 239;
  int n;
  cin >> n;
  unsigned long long s[n];
  int p[n];
  t.assign(n * 4, INF);
  lazy.assign(n * 4, 0);
  for (int i = 0; i < n; i++) {
    cin >> s[i];
  }
  build(s, 0, 0, n - 1);
  for (int i = 1; i <= n; i++) {
    int pos = last_zero(0, 0, n - 1);
    p[pos] = i;
    add(0, 0, n - 1, pos, pos, INF);
    add(0, 0, n - 1, pos + 1, n - 1, -i);
  }
  for (int i = 0; i < n; i++) {
    cout << p[i] << ' ';
  }
}

```