#include <bits/stdc++.h>
using namespace std;
int dx[8] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy[8] = {1, -1, -1, 0, 1, -1, 0, 1};
long long tree[4 * 200005], lazy[4 * 200005], ans[4 * 200005], a[200005];
void build(long long int start, long long int last, long long int node) {
  if (start == last) {
    tree[node] = a[start];
    ans[node] = start;
    return;
  }
  long long int mid = (start + last) >> 1, l = node << 1;
  build(start, mid, l), build(mid + 1, last, l + 1);
  long long t1 = tree[l], t2 = tree[l + 1];
  if (t1 < t2)
    ans[node] = ans[l];
  else
    ans[node] = ans[l + 1];
  tree[node] = (tree[l] > tree[l + 1] ? tree[l + 1] : tree[l]);
  return;
}
void update(long long int start, long long int last, long long int l,
            long long int r, long long int node, long long val) {
  if (lazy[node]) {
    tree[node] += lazy[node];
    if (start != last) {
      lazy[node << 1] += lazy[node];
      lazy[(node << 1) + 1] += lazy[node];
    }
    lazy[node] = 0;
  }
  if (start > r || last < l) return;
  if (l <= start && r >= last) {
    tree[node] += val;
    if (start != last) {
      lazy[node << 1] += val;
      lazy[(node << 1) + 1] += val;
    }
    return;
  }
  long long int mid = (start + last) >> 1;
  update(start, mid, l, r, node << 1, val);
  update(mid + 1, last, l, r, (node << 1) + 1, val);
  if (tree[node << 1] < tree[(node << 1) + 1])
    ans[node] = ans[node << 1];
  else
    ans[node] = ans[(node << 1) + 1];
  tree[node] = (tree[node << 1] > tree[(node << 1) + 1] ? tree[(node << 1) + 1]
                                                        : tree[node << 1]);
  return;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  long long int n1 = n, ans1[n + 1], cnt = 1;
  for (long long int i = 1; i <= n; i++) cin >> a[i];
  build(1, n, 1);
  while (n1--) {
    long long int idx = ans[1];
    ans1[idx] = cnt;
    update(1, n, idx, idx, 1, 1000000000000);
    if (idx < n) update(1, n, idx + 1, n, 1, -cnt);
    cnt++;
  }
  for (long long int i = 1; i <= n; i++) cout << ans1[i] << " ";
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
int dx[8] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy[8] = {1, -1, -1, 0, 1, -1, 0, 1};
long long tree[4 * 200005], lazy[4 * 200005], ans[4 * 200005], a[200005];
void build(long long int start, long long int last, long long int node) {
  if (start == last) {
    tree[node] = a[start];
    ans[node] = start;
    return;
  }
  long long int mid = (start + last) >> 1, l = node << 1;
  build(start, mid, l), build(mid + 1, last, l + 1);
  long long t1 = tree[l], t2 = tree[l + 1];
  if (t1 < t2)
    ans[node] = ans[l];
  else
    ans[node] = ans[l + 1];
  tree[node] = (tree[l] > tree[l + 1] ? tree[l + 1] : tree[l]);
  return;
}
void update(long long int start, long long int last, long long int l,
            long long int r, long long int node, long long val) {
  if (lazy[node]) {
    tree[node] += lazy[node];
    if (start != last) {
      lazy[node << 1] += lazy[node];
      lazy[(node << 1) + 1] += lazy[node];
    }
    lazy[node] = 0;
  }
  if (start > r || last < l) return;
  if (l <= start && r >= last) {
    tree[node] += val;
    if (start != last) {
      lazy[node << 1] += val;
      lazy[(node << 1) + 1] += val;
    }
    return;
  }
  long long int mid = (start + last) >> 1;
  update(start, mid, l, r, node << 1, val);
  update(mid + 1, last, l, r, (node << 1) + 1, val);
  if (tree[node << 1] < tree[(node << 1) + 1])
    ans[node] = ans[node << 1];
  else
    ans[node] = ans[(node << 1) + 1];
  tree[node] = (tree[node << 1] > tree[(node << 1) + 1] ? tree[(node << 1) + 1]
                                                        : tree[node << 1]);
  return;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  long long int n1 = n, ans1[n + 1], cnt = 1;
  for (long long int i = 1; i <= n; i++) cin >> a[i];
  build(1, n, 1);
  while (n1--) {
    long long int idx = ans[1];
    ans1[idx] = cnt;
    update(1, n, idx, idx, 1, 1000000000000);
    if (idx < n) update(1, n, idx + 1, n, 1, -cnt);
    cnt++;
  }
  for (long long int i = 1; i <= n; i++) cout << ans1[i] << " ";
  return 0;
}

```