#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int MAXN = 2e5 + 5;
const int inf = 0x3f3f3f3f;
long long s[MAXN];
struct node {
  int l, r;
  long long sum;
} node[MAXN << 2];
void build(int l, int r, int num) {
  node[num].l = l;
  node[num].r = r;
  if (l == r) {
    node[num].sum = l;
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, num << 1);
  build(mid + 1, r, num << 1 | 1);
  node[num].sum = node[num << 1].sum + node[num << 1 | 1].sum;
}
void updata(int pos, int num) {
  if (node[num].l == node[num].r) {
    node[num].sum = 0;
    return;
  }
  int mid = (node[num].l + node[num].r) >> 1;
  if (pos <= mid)
    updata(pos, num << 1);
  else
    updata(pos, num << 1 | 1);
  node[num].sum = node[num << 1 | 1].sum + node[num << 1].sum;
}
int query(long long val, int num) {
  if (node[num].l == node[num].r) {
    return node[num].l;
  }
  long long sum1 = node[num << 1].sum;
  if (sum1 <= val)
    return query(val - sum1, num << 1 | 1);
  else
    return query(val, num << 1);
}
int dp[MAXN];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
  }
  build(1, n, 1);
  for (int i = n; i >= 1; i--) {
    int ans = query(s[i], 1);
    dp[i] = ans;
    updata(ans, 1);
  }
  for (int i = 1; i <= n; i++) {
    cout << dp[i] << " ";
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 mod = 1e9 + 7;
const int MAXN = 2e5 + 5;
const int inf = 0x3f3f3f3f;
long long s[MAXN];
struct node {
  int l, r;
  long long sum;
} node[MAXN << 2];
void build(int l, int r, int num) {
  node[num].l = l;
  node[num].r = r;
  if (l == r) {
    node[num].sum = l;
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, num << 1);
  build(mid + 1, r, num << 1 | 1);
  node[num].sum = node[num << 1].sum + node[num << 1 | 1].sum;
}
void updata(int pos, int num) {
  if (node[num].l == node[num].r) {
    node[num].sum = 0;
    return;
  }
  int mid = (node[num].l + node[num].r) >> 1;
  if (pos <= mid)
    updata(pos, num << 1);
  else
    updata(pos, num << 1 | 1);
  node[num].sum = node[num << 1 | 1].sum + node[num << 1].sum;
}
int query(long long val, int num) {
  if (node[num].l == node[num].r) {
    return node[num].l;
  }
  long long sum1 = node[num << 1].sum;
  if (sum1 <= val)
    return query(val - sum1, num << 1 | 1);
  else
    return query(val, num << 1);
}
int dp[MAXN];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
  }
  build(1, n, 1);
  for (int i = n; i >= 1; i--) {
    int ans = query(s[i], 1);
    dp[i] = ans;
    updata(ans, 1);
  }
  for (int i = 1; i <= n; i++) {
    cout << dp[i] << " ";
  }
}

```