#include <bits/stdc++.h>
using namespace std;
vector<long long> tree;
vector<long long> s;
vector<long long> p;
long long build(long long L, long long R, long long v) {
  if (R - L == 1) {
    return tree[v] = L + 1;
  } else {
    return tree[v] = build(L, (L + R) / 2, v * 2 + 1) +
                     build((L + R) / 2, R, v * 2 + 2);
  }
}
long long get(long long L, long long R, long long v, long long sum) {
  if (R - L == 1) {
    return L;
  }
  if (tree[v * 2 + 1] <= sum) {
    return get((L + R) / 2, R, v * 2 + 2, sum - tree[v * 2 + 1]);
  } else {
    return get(L, (L + R) / 2, v * 2 + 1, sum);
  }
}
long long sset(long long L, long long R, long long v, long long x) {
  if (R - L == 1) {
    return tree[v] = 0;
  }
  if (x < (L + R) / 2) {
    return tree[v] = sset(L, (L + R) / 2, v * 2 + 1, x) + tree[v * 2 + 2];
  } else {
    return tree[v] = sset((L + R) / 2, R, v * 2 + 2, x) + tree[v * 2 + 1];
  }
}
int32_t main() {
  long long N;
  cin >> N;
  tree.resize(N * 4);
  s.resize(N);
  p.resize(N);
  for (long long i = 0; i < N; i++) {
    cin >> s[i];
  }
  build(0, N, 0);
  for (long long i = N - 1; i >= 0; i--) {
    p[i] = get(0, N, 0, s[i]);
    sset(0, N, 0, p[i]);
    p[i]++;
  }
  for (auto i : p) {
    cout << i << " ";
  }
  cout << endl;
}

### Prompt

Generate 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;
vector<long long> tree;
vector<long long> s;
vector<long long> p;
long long build(long long L, long long R, long long v) {
  if (R - L == 1) {
    return tree[v] = L + 1;
  } else {
    return tree[v] = build(L, (L + R) / 2, v * 2 + 1) +
                     build((L + R) / 2, R, v * 2 + 2);
  }
}
long long get(long long L, long long R, long long v, long long sum) {
  if (R - L == 1) {
    return L;
  }
  if (tree[v * 2 + 1] <= sum) {
    return get((L + R) / 2, R, v * 2 + 2, sum - tree[v * 2 + 1]);
  } else {
    return get(L, (L + R) / 2, v * 2 + 1, sum);
  }
}
long long sset(long long L, long long R, long long v, long long x) {
  if (R - L == 1) {
    return tree[v] = 0;
  }
  if (x < (L + R) / 2) {
    return tree[v] = sset(L, (L + R) / 2, v * 2 + 1, x) + tree[v * 2 + 2];
  } else {
    return tree[v] = sset((L + R) / 2, R, v * 2 + 2, x) + tree[v * 2 + 1];
  }
}
int32_t main() {
  long long N;
  cin >> N;
  tree.resize(N * 4);
  s.resize(N);
  p.resize(N);
  for (long long i = 0; i < N; i++) {
    cin >> s[i];
  }
  build(0, N, 0);
  for (long long i = N - 1; i >= 0; i--) {
    p[i] = get(0, N, 0, s[i]);
    sset(0, N, 0, p[i]);
    p[i]++;
  }
  for (auto i : p) {
    cout << i << " ";
  }
  cout << endl;
}

```