#include <bits/stdc++.h>
using namespace std;
const int LIM = 1e5 + 5, MOD = 1e9 + 7;
const int maxn = 1000006;
long long solve(long long a, long long b) {
  if ((a == b) || (a == 0)) {
    return 0;
  } else {
    long long res;
    res = 1 + solve(a % (b / 2), b / 2);
    return res;
  }
}
long long ft[200005];
long long n;
void update(long long i, long long diff) {
  for (; i <= n; i = i + (i & (-i))) {
    ft[i] += diff;
  }
}
long long query(long long i) {
  long long sum = 0;
  for (; i > 0; i = i - (i & (-i))) {
    sum += ft[i];
  }
  return sum;
}
long long solve(long long z) {
  long long sum = 0, ret = 0, i;
  for (i = 20; i >= 0; i--) {
    if (((ret + (1 << i)) <= n) && ((ft[ret + (1 << i)] + sum) <= z)) {
      ret = ret + (1 << i);
      sum = sum + ft[ret];
    }
  }
  return (ret + 1);
}
int main() {
  long long i, j, k, m;
  cin >> n;
  long long res[n + 5];
  long long s[n + 5];
  for (i = 1; i <= n; i++) ft[i] = 0;
  for (i = 1; i <= n; i++) {
    update(i, i);
    cin >> s[i];
  }
  for (i = n; i >= 1; i--) {
    res[i] = solve(s[i]);
    update(res[i], -res[i]);
  }
  for (i = 1; i <= n; i++) {
    cout << res[i] << " ";
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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 LIM = 1e5 + 5, MOD = 1e9 + 7;
const int maxn = 1000006;
long long solve(long long a, long long b) {
  if ((a == b) || (a == 0)) {
    return 0;
  } else {
    long long res;
    res = 1 + solve(a % (b / 2), b / 2);
    return res;
  }
}
long long ft[200005];
long long n;
void update(long long i, long long diff) {
  for (; i <= n; i = i + (i & (-i))) {
    ft[i] += diff;
  }
}
long long query(long long i) {
  long long sum = 0;
  for (; i > 0; i = i - (i & (-i))) {
    sum += ft[i];
  }
  return sum;
}
long long solve(long long z) {
  long long sum = 0, ret = 0, i;
  for (i = 20; i >= 0; i--) {
    if (((ret + (1 << i)) <= n) && ((ft[ret + (1 << i)] + sum) <= z)) {
      ret = ret + (1 << i);
      sum = sum + ft[ret];
    }
  }
  return (ret + 1);
}
int main() {
  long long i, j, k, m;
  cin >> n;
  long long res[n + 5];
  long long s[n + 5];
  for (i = 1; i <= n; i++) ft[i] = 0;
  for (i = 1; i <= n; i++) {
    update(i, i);
    cin >> s[i];
  }
  for (i = n; i >= 1; i--) {
    res[i] = solve(s[i]);
    update(res[i], -res[i]);
  }
  for (i = 1; i <= n; i++) {
    cout << res[i] << " ";
  }
  return 0;
}

```