#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1000000000")
using namespace std;
const int maxn = (int)2e5 + 10;
int n;
long long v[maxn];
long long t[4 * maxn];
void change(int it, int l, int r, int pos, int x) {
  if (l == r) {
    t[it] = x;
  } else {
    int m = (l + r) >> 1;
    if (pos <= m) {
      change(2 * it, l, m, pos, x);
    } else {
      change(2 * it + 1, m + 1, r, pos, x);
    }
    t[it] = t[2 * it] + t[2 * it + 1];
  }
}
int go(int it, int l, int r, long long s) {
  if (l == r) {
    return l;
  }
  int m = (l + r) >> 1;
  if (t[2 * it] > s) {
    return go(2 * it, l, m, s);
  }
  return go(2 * it + 1, m + 1, r, s - t[2 * it]);
}
int ans[maxn];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &v[i]);
  }
  for (int i = 1; i <= n; i++) {
    change(1, 1, n, i, i);
  }
  for (int i = n - 1; i >= 0; i--) {
    int x = go(1, 1, n, v[i]);
    change(1, 1, n, x, 0);
    ans[i] = x;
  }
  for (int i = 0; i < n; i++) {
    printf("%d ", ans[i]);
  }
  return 0;
}

### Prompt

Please formulate 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:1000000000")
using namespace std;
const int maxn = (int)2e5 + 10;
int n;
long long v[maxn];
long long t[4 * maxn];
void change(int it, int l, int r, int pos, int x) {
  if (l == r) {
    t[it] = x;
  } else {
    int m = (l + r) >> 1;
    if (pos <= m) {
      change(2 * it, l, m, pos, x);
    } else {
      change(2 * it + 1, m + 1, r, pos, x);
    }
    t[it] = t[2 * it] + t[2 * it + 1];
  }
}
int go(int it, int l, int r, long long s) {
  if (l == r) {
    return l;
  }
  int m = (l + r) >> 1;
  if (t[2 * it] > s) {
    return go(2 * it, l, m, s);
  }
  return go(2 * it + 1, m + 1, r, s - t[2 * it]);
}
int ans[maxn];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &v[i]);
  }
  for (int i = 1; i <= n; i++) {
    change(1, 1, n, i, i);
  }
  for (int i = n - 1; i >= 0; i--) {
    int x = go(1, 1, n, v[i]);
    change(1, 1, n, x, 0);
    ans[i] = x;
  }
  for (int i = 0; i < n; i++) {
    printf("%d ", ans[i]);
  }
  return 0;
}

```