#include <bits/stdc++.h>
using namespace std;
const long long N = 3e5;
long long n, t[N], a[N], res[N];
set<long long> st;
void upd(long long i) {
  long long d = i;
  for (i; i < N; i |= i + 1) t[i] += d;
}
long long sum(long long l, long long r) {
  long long s = 0;
  for (; r >= 0; r &= r + 1, r--) s += t[r];
  l--;
  for (; l >= 0; l &= l + 1, l--) s -= t[l];
  return s;
}
long long f(long long x) {
  if (!sum(x, x))
    return (x * x + x) / 2 - sum(0, x) - x;
  else
    return (x * x + x) / 2 - sum(0, x);
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    st.insert(i);
  }
  for (long long i = n - 1; i >= 0; i--) {
    long long l = 1, r = n, med;
    while (r - l > 1) {
      med = (r + l) / 2;
      if (f(med) > a[i])
        r = med;
      else
        l = med;
    }
    if (f(r) == a[i]) {
      res[i] = r;
      upd(r);
    } else {
      res[i] = l;
      upd(l);
    }
  }
  for (long long i = 0; i < n; ++i) cout << res[i] << ' ';
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 long long N = 3e5;
long long n, t[N], a[N], res[N];
set<long long> st;
void upd(long long i) {
  long long d = i;
  for (i; i < N; i |= i + 1) t[i] += d;
}
long long sum(long long l, long long r) {
  long long s = 0;
  for (; r >= 0; r &= r + 1, r--) s += t[r];
  l--;
  for (; l >= 0; l &= l + 1, l--) s -= t[l];
  return s;
}
long long f(long long x) {
  if (!sum(x, x))
    return (x * x + x) / 2 - sum(0, x) - x;
  else
    return (x * x + x) / 2 - sum(0, x);
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    st.insert(i);
  }
  for (long long i = n - 1; i >= 0; i--) {
    long long l = 1, r = n, med;
    while (r - l > 1) {
      med = (r + l) / 2;
      if (f(med) > a[i])
        r = med;
      else
        l = med;
    }
    if (f(r) == a[i]) {
      res[i] = r;
      upd(r);
    } else {
      res[i] = l;
      upd(l);
    }
  }
  for (long long i = 0; i < n; ++i) cout << res[i] << ' ';
}

```