#include <bits/stdc++.h>
using namespace std;
const long long MAX = 200100;
pair<long long, long long> tree[256 * 1024 * 2];
long long lazy[256 * 1024 * 2];
void updateRangeUtil(long long si, long long ss, long long se, long long us,
                     long long ue, long long diff) {
  if (lazy[si] != 0) {
    tree[si].first += lazy[si];
    if (ss != se) {
      lazy[si * 2 + 1] += lazy[si];
      lazy[si * 2 + 2] += lazy[si];
    }
    lazy[si] = 0;
  }
  if (ss > se || ss > ue || se < us) return;
  if (ss >= us && se <= ue) {
    tree[si].first += diff;
    if (ss != se) {
      lazy[si * 2 + 1] += diff;
      lazy[si * 2 + 2] += diff;
    }
    return;
  }
  long long mid = (ss + se) / 2;
  updateRangeUtil(si * 2 + 1, ss, mid, us, ue, diff);
  updateRangeUtil(si * 2 + 2, mid + 1, se, us, ue, diff);
  tree[si] = min(tree[si * 2 + 1], tree[si * 2 + 2]);
}
void updateRange(long long n, long long us, long long ue, long long diff) {
  updateRangeUtil(0, 0, n - 1, us, ue, diff);
}
void updateRangeUtil2(long long si, long long ss, long long se, long long us,
                      long long ue, long long diff) {
  if (ss > se || ss > ue || se < us) return;
  if (ss >= us && se <= ue) {
    tree[si].second += diff;
    return;
  }
  long long mid = (ss + se) / 2;
  updateRangeUtil2(si * 2 + 1, ss, mid, us, ue, diff);
  updateRangeUtil2(si * 2 + 2, mid + 1, se, us, ue, diff);
  tree[si] = min(tree[si * 2 + 1], tree[si * 2 + 2]);
}
void updateRange2(long long n, long long us, long long ue, long long diff) {
  updateRangeUtil2(0, 0, n - 1, us, ue, diff);
}
int main() {
  long long arr[MAX];
  long long n;
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    cin >> arr[i];
  }
  reverse(arr, arr + n);
  for (long long i = 0; i < n; ++i) {
    updateRange2(n, i, i, i);
  }
  for (long long i = 0; i < n; ++i) {
    updateRange(n, i, i, arr[i]);
  }
  long long res[MAX];
  for (long long iter = 1; iter <= n; ++iter) {
    long long smallest = tree[0].second;
    updateRange(n, 0, smallest, -iter);
    updateRange(n, smallest, smallest, 1LL << 62);
    res[smallest] = iter;
  }
  reverse(res, res + n);
  for (long long i = 0; i < n; ++i) {
    if (i) cout << ' ';
    cout << res[i];
  }
  cout << endl;
}

### Prompt

Please create a solution in Cpp 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;
const long long MAX = 200100;
pair<long long, long long> tree[256 * 1024 * 2];
long long lazy[256 * 1024 * 2];
void updateRangeUtil(long long si, long long ss, long long se, long long us,
                     long long ue, long long diff) {
  if (lazy[si] != 0) {
    tree[si].first += lazy[si];
    if (ss != se) {
      lazy[si * 2 + 1] += lazy[si];
      lazy[si * 2 + 2] += lazy[si];
    }
    lazy[si] = 0;
  }
  if (ss > se || ss > ue || se < us) return;
  if (ss >= us && se <= ue) {
    tree[si].first += diff;
    if (ss != se) {
      lazy[si * 2 + 1] += diff;
      lazy[si * 2 + 2] += diff;
    }
    return;
  }
  long long mid = (ss + se) / 2;
  updateRangeUtil(si * 2 + 1, ss, mid, us, ue, diff);
  updateRangeUtil(si * 2 + 2, mid + 1, se, us, ue, diff);
  tree[si] = min(tree[si * 2 + 1], tree[si * 2 + 2]);
}
void updateRange(long long n, long long us, long long ue, long long diff) {
  updateRangeUtil(0, 0, n - 1, us, ue, diff);
}
void updateRangeUtil2(long long si, long long ss, long long se, long long us,
                      long long ue, long long diff) {
  if (ss > se || ss > ue || se < us) return;
  if (ss >= us && se <= ue) {
    tree[si].second += diff;
    return;
  }
  long long mid = (ss + se) / 2;
  updateRangeUtil2(si * 2 + 1, ss, mid, us, ue, diff);
  updateRangeUtil2(si * 2 + 2, mid + 1, se, us, ue, diff);
  tree[si] = min(tree[si * 2 + 1], tree[si * 2 + 2]);
}
void updateRange2(long long n, long long us, long long ue, long long diff) {
  updateRangeUtil2(0, 0, n - 1, us, ue, diff);
}
int main() {
  long long arr[MAX];
  long long n;
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    cin >> arr[i];
  }
  reverse(arr, arr + n);
  for (long long i = 0; i < n; ++i) {
    updateRange2(n, i, i, i);
  }
  for (long long i = 0; i < n; ++i) {
    updateRange(n, i, i, arr[i]);
  }
  long long res[MAX];
  for (long long iter = 1; iter <= n; ++iter) {
    long long smallest = tree[0].second;
    updateRange(n, 0, smallest, -iter);
    updateRange(n, smallest, smallest, 1LL << 62);
    res[smallest] = iter;
  }
  reverse(res, res + n);
  for (long long i = 0; i < n; ++i) {
    if (i) cout << ' ';
    cout << res[i];
  }
  cout << endl;
}

```