#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:640000000")
using namespace std;
const double eps = 1e-9;
const double pi = acos(-1.0);
const int maxn = (int)2e5 + 10;
long long a[maxn];
int res[maxn];
long long t[4 * maxn];
void build(int idx, int l, int r) {
  if (l == r) {
    t[idx] = l;
  } else {
    int m = (l + r) >> 1;
    build(2 * idx, l, m);
    build(2 * idx + 1, m + 1, r);
    t[idx] = t[2 * idx] + t[2 * idx + 1];
  }
}
int get(int idx, int l, int r, long long sum) {
  if (l == r) return l;
  int m = (l + r) >> 1;
  if (t[2 * idx] > sum) {
    return get(2 * idx, l, m, sum);
  }
  return get(2 * idx + 1, m + 1, r, sum - t[2 * idx]);
}
void upd(int idx, int l, int r, int pos) {
  if (l == r) {
    t[idx] = 0;
  } else {
    int m = (l + r) >> 1;
    if (pos <= m) {
      upd(2 * idx, l, m, pos);
    } else {
      upd(2 * idx + 1, m + 1, r, pos);
    }
    t[idx] = t[2 * idx] + t[2 * idx + 1];
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%I64d", &a[i]);
  build(1, 1, n);
  for (int i = n - 1; i >= 0; i--) {
    int val = get(1, 1, n, a[i]);
    res[i] = val;
    upd(1, 1, n, val);
  }
  for (int i = 0; i < n; i++) {
    printf("%d ", res[i]);
  }
  return 0;
}

### 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>
#pragma comment(linker, "/STACK:640000000")
using namespace std;
const double eps = 1e-9;
const double pi = acos(-1.0);
const int maxn = (int)2e5 + 10;
long long a[maxn];
int res[maxn];
long long t[4 * maxn];
void build(int idx, int l, int r) {
  if (l == r) {
    t[idx] = l;
  } else {
    int m = (l + r) >> 1;
    build(2 * idx, l, m);
    build(2 * idx + 1, m + 1, r);
    t[idx] = t[2 * idx] + t[2 * idx + 1];
  }
}
int get(int idx, int l, int r, long long sum) {
  if (l == r) return l;
  int m = (l + r) >> 1;
  if (t[2 * idx] > sum) {
    return get(2 * idx, l, m, sum);
  }
  return get(2 * idx + 1, m + 1, r, sum - t[2 * idx]);
}
void upd(int idx, int l, int r, int pos) {
  if (l == r) {
    t[idx] = 0;
  } else {
    int m = (l + r) >> 1;
    if (pos <= m) {
      upd(2 * idx, l, m, pos);
    } else {
      upd(2 * idx + 1, m + 1, r, pos);
    }
    t[idx] = t[2 * idx] + t[2 * idx + 1];
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%I64d", &a[i]);
  build(1, 1, n);
  for (int i = n - 1; i >= 0; i--) {
    int val = get(1, 1, n, a[i]);
    res[i] = val;
    upd(1, 1, n, val);
  }
  for (int i = 0; i < n; i++) {
    printf("%d ", res[i]);
  }
  return 0;
}

```