#include <bits/stdc++.h>
using namespace std;
long long bit[200005];
void add(long long idx, long long val) {
  for (int i = idx; i < 200005; i += (i & (-i))) {
    bit[i] += val;
  }
}
long long query(long long i) {
  long long ans = 0;
  for (; i > 0; i -= (i & (-i))) ans += bit[i];
  return ans;
}
long long b[200005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, i, j, k, m;
  cin >> n;
  long long a[n];
  for (i = 0; i < n; i++) cin >> a[i];
  set<long long> s, s1;
  unordered_map<long long, long long> mp;
  long long sm = 0;
  long long sum[n + 5];
  for (i = 1; i <= n; i++) {
    sm += i;
    mp[sm] = i;
    s.insert(i);
    s1.insert(sm);
    sum[i] = sm;
  }
  for (i = 1; i <= n; i++) {
    add(i, i);
  }
  for (i = n - 1; i >= 0; i--) {
    long long f = 1, l = n;
    long long mid;
    long long anss = 0;
    while (f <= l) {
      mid = (f + l) / 2;
      if (query(mid) > a[i]) {
        l = mid - 1;
      } else {
        f = mid + 1;
        anss = mid;
      }
    }
    anss++;
    b[i] = anss;
    add(anss, -anss);
  }
  for (i = 0; i < n; i++) cout << b[i] << " ";
}

### Prompt

Your challenge is to write 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;
long long bit[200005];
void add(long long idx, long long val) {
  for (int i = idx; i < 200005; i += (i & (-i))) {
    bit[i] += val;
  }
}
long long query(long long i) {
  long long ans = 0;
  for (; i > 0; i -= (i & (-i))) ans += bit[i];
  return ans;
}
long long b[200005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, i, j, k, m;
  cin >> n;
  long long a[n];
  for (i = 0; i < n; i++) cin >> a[i];
  set<long long> s, s1;
  unordered_map<long long, long long> mp;
  long long sm = 0;
  long long sum[n + 5];
  for (i = 1; i <= n; i++) {
    sm += i;
    mp[sm] = i;
    s.insert(i);
    s1.insert(sm);
    sum[i] = sm;
  }
  for (i = 1; i <= n; i++) {
    add(i, i);
  }
  for (i = n - 1; i >= 0; i--) {
    long long f = 1, l = n;
    long long mid;
    long long anss = 0;
    while (f <= l) {
      mid = (f + l) / 2;
      if (query(mid) > a[i]) {
        l = mid - 1;
      } else {
        f = mid + 1;
        anss = mid;
      }
    }
    anss++;
    b[i] = anss;
    add(anss, -anss);
  }
  for (i = 0; i < n; i++) cout << b[i] << " ";
}

```