#include <bits/stdc++.h>
using namespace std;
const long double EPS = 1e-9, PI = acos(-1.);
const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int N = 2e5 + 5;
int n;
long long mi[4 * N], pos[4 * N], v[N], lz[4 * N];
void build(int p, int l, int r) {
  if (l == r) {
    mi[p] = v[l], pos[p] = l;
    return;
  }
  build(2 * p, l, (l + r) / 2);
  build(2 * p + 1, (l + r) / 2 + 1, r);
  mi[p] = min(mi[2 * p], mi[2 * p + 1]);
  if (mi[p] == mi[2 * p + 1])
    pos[p] = pos[2 * p + 1];
  else
    pos[p] = pos[2 * p];
}
void push(int p, int l, int r) {
  if (lz[p]) {
    mi[p] += lz[p];
    if (l != r) {
      lz[2 * p] += lz[p];
      lz[2 * p + 1] += lz[p];
    }
    lz[p] = 0;
  }
}
pair<long long, long long> query(int p, int l, int r, int i, int j) {
  push(p, l, r);
  if (r < i or l > j) return {LINF, 0};
  if (l >= i and r <= j) return {mi[p], pos[p]};
  pair<long long, long long> a = query(2 * p, l, (l + r) / 2, i, j);
  pair<long long, long long> b = query(2 * p + 1, (l + r) / 2 + 1, r, i, j);
  if (a.first < b.first)
    return a;
  else
    return b;
}
void update(int p, int l, int r, int i, int j, long long v) {
  push(p, l, r);
  if (r < i or l > j) return;
  if (l >= i and r <= j) {
    lz[p] = v;
    push(p, l, r);
    return;
  }
  update(2 * p, l, (l + r) / 2, i, j, v);
  update(2 * p + 1, (l + r) / 2 + 1, r, i, j, v);
  mi[p] = min(mi[2 * p], mi[2 * p + 1]);
  if (mi[p] == mi[2 * p + 1])
    pos[p] = pos[2 * p + 1];
  else
    pos[p] = pos[2 * p];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &v[i]);
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    auto p = query(1, 1, n, 1, n);
    assert(p.second);
    v[p.second] = i;
    update(1, 1, n, p.second + 1, n, -i);
    update(1, 1, n, p.second, p.second, LINF);
  }
  for (int i = 1; i <= n; i++) printf("%lld ", v[i]);
  printf("\n");
  return 0;
}

### 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;
const long double EPS = 1e-9, PI = acos(-1.);
const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int N = 2e5 + 5;
int n;
long long mi[4 * N], pos[4 * N], v[N], lz[4 * N];
void build(int p, int l, int r) {
  if (l == r) {
    mi[p] = v[l], pos[p] = l;
    return;
  }
  build(2 * p, l, (l + r) / 2);
  build(2 * p + 1, (l + r) / 2 + 1, r);
  mi[p] = min(mi[2 * p], mi[2 * p + 1]);
  if (mi[p] == mi[2 * p + 1])
    pos[p] = pos[2 * p + 1];
  else
    pos[p] = pos[2 * p];
}
void push(int p, int l, int r) {
  if (lz[p]) {
    mi[p] += lz[p];
    if (l != r) {
      lz[2 * p] += lz[p];
      lz[2 * p + 1] += lz[p];
    }
    lz[p] = 0;
  }
}
pair<long long, long long> query(int p, int l, int r, int i, int j) {
  push(p, l, r);
  if (r < i or l > j) return {LINF, 0};
  if (l >= i and r <= j) return {mi[p], pos[p]};
  pair<long long, long long> a = query(2 * p, l, (l + r) / 2, i, j);
  pair<long long, long long> b = query(2 * p + 1, (l + r) / 2 + 1, r, i, j);
  if (a.first < b.first)
    return a;
  else
    return b;
}
void update(int p, int l, int r, int i, int j, long long v) {
  push(p, l, r);
  if (r < i or l > j) return;
  if (l >= i and r <= j) {
    lz[p] = v;
    push(p, l, r);
    return;
  }
  update(2 * p, l, (l + r) / 2, i, j, v);
  update(2 * p + 1, (l + r) / 2 + 1, r, i, j, v);
  mi[p] = min(mi[2 * p], mi[2 * p + 1]);
  if (mi[p] == mi[2 * p + 1])
    pos[p] = pos[2 * p + 1];
  else
    pos[p] = pos[2 * p];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &v[i]);
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    auto p = query(1, 1, n, 1, n);
    assert(p.second);
    v[p.second] = i;
    update(1, 1, n, p.second + 1, n, -i);
    update(1, 1, n, p.second, p.second, LINF);
  }
  for (int i = 1; i <= n; i++) printf("%lld ", v[i]);
  printf("\n");
  return 0;
}

```