#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
long long a[N];
pair<long long, long long> seg[4 * N];
void build(long long ind, long long l, long long r) {
  if (l > r) return;
  if (l == r) {
    seg[ind] = {a[l], l};
    return;
  }
  long long mid = (l + r) / 2;
  build(2 * ind + 1, l, mid);
  build(2 * ind + 2, mid + 1, r);
  if (seg[2 * ind + 1].first < seg[2 * ind + 2].first)
    seg[ind] = seg[2 * ind + 1];
  else
    seg[ind] = seg[2 * ind + 2];
}
long long lazy[4 * N];
void update(long long ind, long long l, long long r, long long lq, long long rq,
            long long val) {
  if (lazy[ind] != 0) {
    seg[ind].first += lazy[ind];
    if (l != r) {
      lazy[ind * 2 + 1] += lazy[ind];
      lazy[ind * 2 + 2] += lazy[ind];
    }
    lazy[ind] = 0;
  }
  if (l > r || r < lq || rq < l) return;
  if (l >= lq && r <= rq) {
    seg[ind].first += val;
    if (l != r) {
      lazy[ind * 2 + 1] += val;
      lazy[ind * 2 + 2] += val;
    }
    return;
  }
  long long mid = (l + r) / 2;
  update(2 * ind + 1, l, mid, lq, rq, val);
  update(2 * ind + 2, mid + 1, r, lq, rq, val);
  if (seg[2 * ind + 1].first < seg[2 * ind + 2].first)
    seg[ind] = seg[2 * ind + 1];
  else
    seg[ind] = seg[2 * ind + 2];
}
pair<long long, long long> query(long long ind, long long l, long long r,
                                 long long lq, long long rq) {
  if (l > r || r < lq || rq < l) return {1000000007, 0};
  if (lazy[ind] != 0) {
    seg[ind].first += lazy[ind];
    if (l != r) {
      lazy[ind * 2 + 1] += lazy[ind];
      lazy[ind * 2 + 2] += lazy[ind];
    }
    lazy[ind] = 0;
  }
  if (l >= lq && r <= rq) return seg[ind];
  long long mid = (l + r) / 2;
  pair<long long, long long> k1, k2;
  k1 = query(2 * ind + 1, l, mid, lq, rq);
  k2 = query(2 * ind + 2, mid + 1, r, lq, rq);
  if (k1.first < k2.first)
    return k1;
  else
    return k2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  long long n, i, k;
  cin >> n;
  for (i = 0; i < n; i++) cin >> a[i];
  build(0, 0, n - 1);
  long long ans[n];
  pair<long long, long long> x;
  long long l;
  k = 1;
  for (i = 0; i < n; i++) {
    x = query(0, 0, n - 1, 0, n - 1);
    l = x.second;
    ans[l] = k;
    update(0, 0, n - 1, l, l, 1000000007000000);
    update(0, 0, n - 1, l + 1, n - 1, -k);
    k++;
  }
  for (i = 0; i < n; i++) cout << ans[i] << " ";
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
const int N = 200005;
long long a[N];
pair<long long, long long> seg[4 * N];
void build(long long ind, long long l, long long r) {
  if (l > r) return;
  if (l == r) {
    seg[ind] = {a[l], l};
    return;
  }
  long long mid = (l + r) / 2;
  build(2 * ind + 1, l, mid);
  build(2 * ind + 2, mid + 1, r);
  if (seg[2 * ind + 1].first < seg[2 * ind + 2].first)
    seg[ind] = seg[2 * ind + 1];
  else
    seg[ind] = seg[2 * ind + 2];
}
long long lazy[4 * N];
void update(long long ind, long long l, long long r, long long lq, long long rq,
            long long val) {
  if (lazy[ind] != 0) {
    seg[ind].first += lazy[ind];
    if (l != r) {
      lazy[ind * 2 + 1] += lazy[ind];
      lazy[ind * 2 + 2] += lazy[ind];
    }
    lazy[ind] = 0;
  }
  if (l > r || r < lq || rq < l) return;
  if (l >= lq && r <= rq) {
    seg[ind].first += val;
    if (l != r) {
      lazy[ind * 2 + 1] += val;
      lazy[ind * 2 + 2] += val;
    }
    return;
  }
  long long mid = (l + r) / 2;
  update(2 * ind + 1, l, mid, lq, rq, val);
  update(2 * ind + 2, mid + 1, r, lq, rq, val);
  if (seg[2 * ind + 1].first < seg[2 * ind + 2].first)
    seg[ind] = seg[2 * ind + 1];
  else
    seg[ind] = seg[2 * ind + 2];
}
pair<long long, long long> query(long long ind, long long l, long long r,
                                 long long lq, long long rq) {
  if (l > r || r < lq || rq < l) return {1000000007, 0};
  if (lazy[ind] != 0) {
    seg[ind].first += lazy[ind];
    if (l != r) {
      lazy[ind * 2 + 1] += lazy[ind];
      lazy[ind * 2 + 2] += lazy[ind];
    }
    lazy[ind] = 0;
  }
  if (l >= lq && r <= rq) return seg[ind];
  long long mid = (l + r) / 2;
  pair<long long, long long> k1, k2;
  k1 = query(2 * ind + 1, l, mid, lq, rq);
  k2 = query(2 * ind + 2, mid + 1, r, lq, rq);
  if (k1.first < k2.first)
    return k1;
  else
    return k2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  long long n, i, k;
  cin >> n;
  for (i = 0; i < n; i++) cin >> a[i];
  build(0, 0, n - 1);
  long long ans[n];
  pair<long long, long long> x;
  long long l;
  k = 1;
  for (i = 0; i < n; i++) {
    x = query(0, 0, n - 1, 0, n - 1);
    l = x.second;
    ans[l] = k;
    update(0, 0, n - 1, l, l, 1000000007000000);
    update(0, 0, n - 1, l + 1, n - 1, -k);
    k++;
  }
  for (i = 0; i < n; i++) cout << ans[i] << " ";
  return 0;
}

```