#include <bits/stdc++.h>
using namespace std;
const long long Nmax = 2e6 + 5;
long long v[Nmax];
long long ans[Nmax];
long long lazy[4 * Nmax];
pair<long long, long long> all[4 * Nmax];
long long p, q, val;
void add(long long &first, long long val) {
  if (first != LLONG_MAX) first += val;
}
void propagate(long long node) {
  if (lazy[node]) {
    add(lazy[(node << 1)], lazy[node]);
    add(lazy[((node << 1) + 1)], lazy[node]);
    add(all[(node << 1)].first, lazy[node]);
    add(all[((node << 1) + 1)].first, lazy[node]);
    lazy[node] = 0;
  }
}
void build(long long lo, long long hi, long long node) {
  if (lo == hi) {
    all[node] = make_pair(v[lo], -lo);
  } else {
    long long mid = (lo + hi) >> 1;
    build(lo, mid, (node << 1));
    build(mid + 1, hi, ((node << 1) + 1));
    all[node] = min(all[(node << 1)], all[((node << 1) + 1)]);
  }
}
void update(long long lo, long long hi, long long node) {
  propagate(node);
  if (p <= lo and hi <= q) {
    lazy[node] += val;
    all[node].first += val;
  } else {
    long long mid = (lo + hi) >> 1;
    if (p <= mid) update(lo, mid, (node << 1));
    if (mid < q) update(mid + 1, hi, ((node << 1) + 1));
    all[node] = min(all[(node << 1)], all[((node << 1) + 1)]);
  }
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; ++i) {
    cin >> v[i];
  }
  build(1, n, 1);
  for (long long pos, i = 1; i <= n; ++i) {
    pos = -all[1].second;
    ans[pos] = i;
    p = q = pos;
    val = LLONG_MAX;
    update(1, n, 1);
    p = pos + 1;
    q = n;
    val = -i;
    if (p <= q) update(1, n, 1);
  }
  for (long long i = 1; i <= n; ++i) cout << ans[i] << ' ';
  return 0;
}

### Prompt

Create a solution in Cpp for 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 Nmax = 2e6 + 5;
long long v[Nmax];
long long ans[Nmax];
long long lazy[4 * Nmax];
pair<long long, long long> all[4 * Nmax];
long long p, q, val;
void add(long long &first, long long val) {
  if (first != LLONG_MAX) first += val;
}
void propagate(long long node) {
  if (lazy[node]) {
    add(lazy[(node << 1)], lazy[node]);
    add(lazy[((node << 1) + 1)], lazy[node]);
    add(all[(node << 1)].first, lazy[node]);
    add(all[((node << 1) + 1)].first, lazy[node]);
    lazy[node] = 0;
  }
}
void build(long long lo, long long hi, long long node) {
  if (lo == hi) {
    all[node] = make_pair(v[lo], -lo);
  } else {
    long long mid = (lo + hi) >> 1;
    build(lo, mid, (node << 1));
    build(mid + 1, hi, ((node << 1) + 1));
    all[node] = min(all[(node << 1)], all[((node << 1) + 1)]);
  }
}
void update(long long lo, long long hi, long long node) {
  propagate(node);
  if (p <= lo and hi <= q) {
    lazy[node] += val;
    all[node].first += val;
  } else {
    long long mid = (lo + hi) >> 1;
    if (p <= mid) update(lo, mid, (node << 1));
    if (mid < q) update(mid + 1, hi, ((node << 1) + 1));
    all[node] = min(all[(node << 1)], all[((node << 1) + 1)]);
  }
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; ++i) {
    cin >> v[i];
  }
  build(1, n, 1);
  for (long long pos, i = 1; i <= n; ++i) {
    pos = -all[1].second;
    ans[pos] = i;
    p = q = pos;
    val = LLONG_MAX;
    update(1, n, 1);
    p = pos + 1;
    q = n;
    val = -i;
    if (p <= q) update(1, n, 1);
  }
  for (long long i = 1; i <= n; ++i) cout << ans[i] << ' ';
  return 0;
}

```