#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
const int MAXN = 2e5 + 20;
const int SIZE = (1 << 19) + 20;
long long s[MAXN];
int ans[MAXN];
struct Segment_tree {
  struct Node {
    int sl, sr;
    long long val;
  } tree[SIZE];
  inline void update(int root) {
    tree[root].val = tree[(root << 1)].val + tree[(root << 1 | 1)].val;
  }
  inline void build(int root, int l, int r) {
    tree[root].sl = l;
    tree[root].sr = r;
    if (l == r) {
      tree[root].val = l;
      return;
    }
    build((root << 1), l, ((tree[root].sl + tree[root].sr) >> 1));
    build((root << 1 | 1), ((tree[root].sl + tree[root].sr) >> 1) + 1, r);
    update(root);
  }
  inline int find(int root, long long v) {
    if (tree[root].sl == tree[root].sr) {
      tree[root].val = 0;
      return tree[root].sl;
    }
    int res = 0;
    if (tree[(root << 1)].val > v) {
      res = find((root << 1), v);
    } else {
      res = find((root << 1 | 1), v - tree[(root << 1)].val);
    }
    update(root);
    return res;
  }
} Tree;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  int n;
  cin >> n;
  for (register int i = (1); i <= (n); ++i) {
    cin >> s[i];
  }
  Tree.build(1, 1, n);
  for (register int i = (n); i >= (1); --i) {
    ans[i] = Tree.find(1, s[i]);
  }
  for (register int i = (1); i <= (n); ++i) {
    cout << ans[i] << ' ';
  }
  cout << '\n';
  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;
#pragma GCC optimize("Ofast")
const int MAXN = 2e5 + 20;
const int SIZE = (1 << 19) + 20;
long long s[MAXN];
int ans[MAXN];
struct Segment_tree {
  struct Node {
    int sl, sr;
    long long val;
  } tree[SIZE];
  inline void update(int root) {
    tree[root].val = tree[(root << 1)].val + tree[(root << 1 | 1)].val;
  }
  inline void build(int root, int l, int r) {
    tree[root].sl = l;
    tree[root].sr = r;
    if (l == r) {
      tree[root].val = l;
      return;
    }
    build((root << 1), l, ((tree[root].sl + tree[root].sr) >> 1));
    build((root << 1 | 1), ((tree[root].sl + tree[root].sr) >> 1) + 1, r);
    update(root);
  }
  inline int find(int root, long long v) {
    if (tree[root].sl == tree[root].sr) {
      tree[root].val = 0;
      return tree[root].sl;
    }
    int res = 0;
    if (tree[(root << 1)].val > v) {
      res = find((root << 1), v);
    } else {
      res = find((root << 1 | 1), v - tree[(root << 1)].val);
    }
    update(root);
    return res;
  }
} Tree;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  int n;
  cin >> n;
  for (register int i = (1); i <= (n); ++i) {
    cin >> s[i];
  }
  Tree.build(1, 1, n);
  for (register int i = (n); i >= (1); --i) {
    ans[i] = Tree.find(1, s[i]);
  }
  for (register int i = (1); i <= (n); ++i) {
    cout << ans[i] << ' ';
  }
  cout << '\n';
  return 0;
}

```