#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18;
const double EPS = 1e-9;
const long long MOD = (long long)(1e9 + 7);
const long long MAXV = (long long)(2e5 + 10);
const long long MAXE = (long long)(1e6 + 10);
long long seg[MAXV * 4];
long long lazy[MAXV * 4];
long long A[MAXV * 4];
void build(int n, int s, int e) {
  if (s + 1 == e) {
    seg[n] = A[s];
  } else {
    int m = (s + e) >> 1;
    build(n * 2, s, m);
    build(n * 2 + 1, m, e);
    seg[n] = min(seg[n * 2], seg[n * 2 + 1]);
  }
}
void add_lazy(int n, long long x) {
  lazy[n] += x;
  seg[n] += x;
}
void push(int n, int s, int e) {
  if (s + 1 == e) return;
  add_lazy(n * 2, lazy[n]);
  add_lazy(n * 2 + 1, lazy[n]);
  lazy[n] = 0;
}
void update(int n, int s, int e, int a, int b, long long x) {
  if (s >= b || e <= a) return;
  if (s >= a && e <= b)
    add_lazy(n, x);
  else {
    if (lazy[n]) push(n, s, e);
    int m = (s + e) >> 1;
    update(n * 2, s, m, a, b, x);
    update(n * 2 + 1, m, e, a, b, x);
    seg[n] = min(seg[n * 2], seg[n * 2 + 1]);
  }
}
int query(int n, int s, int e) {
  if (s + 1 == e)
    return s;
  else {
    if (lazy[n]) push(n, s, e);
    int m = (s + e) >> 1;
    long long ans;
    if (seg[n * 2] < seg[n * 2 + 1])
      ans = query(n * 2, s, m);
    else
      ans = query(n * 2 + 1, m, e);
    seg[n] = min(seg[n * 2], seg[n * 2 + 1]);
    return ans;
  }
}
int n;
long long ans[MAXV];
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> A[i];
  build(1, 0, n);
  long long curr = 1;
  for (int i = 0; i < n; i++) {
    int p = query(1, 0, n);
    ans[p] = curr;
    update(1, 0, n, p + 1, n, -curr);
    update(1, 0, n, p, p + 1, INF);
    curr++;
  }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << "\n";
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 INF = 1e18;
const double EPS = 1e-9;
const long long MOD = (long long)(1e9 + 7);
const long long MAXV = (long long)(2e5 + 10);
const long long MAXE = (long long)(1e6 + 10);
long long seg[MAXV * 4];
long long lazy[MAXV * 4];
long long A[MAXV * 4];
void build(int n, int s, int e) {
  if (s + 1 == e) {
    seg[n] = A[s];
  } else {
    int m = (s + e) >> 1;
    build(n * 2, s, m);
    build(n * 2 + 1, m, e);
    seg[n] = min(seg[n * 2], seg[n * 2 + 1]);
  }
}
void add_lazy(int n, long long x) {
  lazy[n] += x;
  seg[n] += x;
}
void push(int n, int s, int e) {
  if (s + 1 == e) return;
  add_lazy(n * 2, lazy[n]);
  add_lazy(n * 2 + 1, lazy[n]);
  lazy[n] = 0;
}
void update(int n, int s, int e, int a, int b, long long x) {
  if (s >= b || e <= a) return;
  if (s >= a && e <= b)
    add_lazy(n, x);
  else {
    if (lazy[n]) push(n, s, e);
    int m = (s + e) >> 1;
    update(n * 2, s, m, a, b, x);
    update(n * 2 + 1, m, e, a, b, x);
    seg[n] = min(seg[n * 2], seg[n * 2 + 1]);
  }
}
int query(int n, int s, int e) {
  if (s + 1 == e)
    return s;
  else {
    if (lazy[n]) push(n, s, e);
    int m = (s + e) >> 1;
    long long ans;
    if (seg[n * 2] < seg[n * 2 + 1])
      ans = query(n * 2, s, m);
    else
      ans = query(n * 2 + 1, m, e);
    seg[n] = min(seg[n * 2], seg[n * 2 + 1]);
    return ans;
  }
}
int n;
long long ans[MAXV];
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> A[i];
  build(1, 0, n);
  long long curr = 1;
  for (int i = 0; i < n; i++) {
    int p = query(1, 0, n);
    ans[p] = curr;
    update(1, 0, n, p + 1, n, -curr);
    update(1, 0, n, p, p + 1, INF);
    curr++;
  }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << "\n";
  return 0;
}

```