#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
long long BIT[N], s[N];
int n;
int ans[N];
void update(int x, int delta) {
  for (; x <= n; x += x & -x) BIT[x] += delta;
}
long long query(int x) {
  long long sum = 0;
  for (; x > 0; x -= x & -x) sum += BIT[x];
  return sum;
}
int searchNumber(long long prefSum) {
  int num = 0;
  long long sum = 0;
  for (int i = 21; i >= 0; --i) {
    if ((num + (1 << i) <= n) && (sum + BIT[num + (1 << i)] <= prefSum)) {
      num += (1 << i);
      sum += BIT[num];
    }
  }
  return num + 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    update(i, i);
    scanf("%I64d", &s[i]);
  }
  for (int i = n; i >= 1; --i) {
    ans[i] = searchNumber(s[i]);
    update(ans[i], -ans[i]);
  }
  for (int i = 1; i <= n; ++i) {
    printf("%d", ans[i]);
    if (i < n) {
      printf(" ");
    } else
      printf("\n");
  }
}

### 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>
using namespace std;
const int N = 2e5 + 5;
long long BIT[N], s[N];
int n;
int ans[N];
void update(int x, int delta) {
  for (; x <= n; x += x & -x) BIT[x] += delta;
}
long long query(int x) {
  long long sum = 0;
  for (; x > 0; x -= x & -x) sum += BIT[x];
  return sum;
}
int searchNumber(long long prefSum) {
  int num = 0;
  long long sum = 0;
  for (int i = 21; i >= 0; --i) {
    if ((num + (1 << i) <= n) && (sum + BIT[num + (1 << i)] <= prefSum)) {
      num += (1 << i);
      sum += BIT[num];
    }
  }
  return num + 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    update(i, i);
    scanf("%I64d", &s[i]);
  }
  for (int i = n; i >= 1; --i) {
    ans[i] = searchNumber(s[i]);
    update(ans[i], -ans[i]);
  }
  for (int i = 1; i <= n; ++i) {
    printf("%d", ans[i]);
    if (i < n) {
      printf(" ");
    } else
      printf("\n");
  }
}

```