#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b) {
  long long res = 1;
  a = a % 1000000007;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
      b--;
    }
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long fermat_inv(long long y) { return power(y, 1000000007 - 2); }
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
long long sum[800050];
void build(long long a[], int v, int l, int r) {
  if (l == r)
    sum[v] = a[l];
  else {
    int m = (l + r) / 2;
    build(a, 2 * v, l, m);
    build(a, 2 * v + 1, m + 1, r);
    sum[v] = sum[2 * v] + sum[2 * v + 1];
  }
}
long long query(int v, int tl, int tr, int l, int r) {
  if (l > r)
    return 0;
  else if (l == tl && r == tr)
    return sum[v];
  else {
    int m = (tl + tr) / 2;
    return query(2 * v, tl, m, l, min(r, m)) +
           query(2 * v + 1, m + 1, tr, max(l, m + 1), r);
  }
}
void update(int v, int l, int r, int pos, int val) {
  if (l == r)
    sum[v] = val;
  else {
    int m = (l + r) / 2;
    if (pos <= m)
      update(2 * v, l, m, pos, val);
    else
      update(2 * v + 1, m + 1, r, pos, val);
    sum[v] = sum[2 * v] + sum[2 * v + 1];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    int n;
    cin >> n;
    long long b[n];
    long long a[n + 1];
    a[0] = 0;
    for (int i = 0; i < n; i++) {
      cin >> b[i];
      a[i + 1] = i + 1;
    }
    build(a, 1, 0, n);
    for (int i = n - 1; i >= 0; i--) {
      long long l = 0;
      long long r = n;
      long long pos = -1;
      while (l <= r) {
        long long m = (l + r) / 2;
        long long val = query(1, 0, n, 0, m);
        if (val <= b[i]) {
          pos = m;
          l = m + 1;
        } else
          r = m - 1;
      }
      b[i] = pos + 1;
      update(1, 0, n, pos + 1, 0);
    }
    for (int i = 0; i < n; i++) {
      cout << b[i] << " ";
    }
    cout << "\n";
  }
  return 0;
}

### 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;
long long power(long long a, long long b) {
  long long res = 1;
  a = a % 1000000007;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
      b--;
    }
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long fermat_inv(long long y) { return power(y, 1000000007 - 2); }
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
long long sum[800050];
void build(long long a[], int v, int l, int r) {
  if (l == r)
    sum[v] = a[l];
  else {
    int m = (l + r) / 2;
    build(a, 2 * v, l, m);
    build(a, 2 * v + 1, m + 1, r);
    sum[v] = sum[2 * v] + sum[2 * v + 1];
  }
}
long long query(int v, int tl, int tr, int l, int r) {
  if (l > r)
    return 0;
  else if (l == tl && r == tr)
    return sum[v];
  else {
    int m = (tl + tr) / 2;
    return query(2 * v, tl, m, l, min(r, m)) +
           query(2 * v + 1, m + 1, tr, max(l, m + 1), r);
  }
}
void update(int v, int l, int r, int pos, int val) {
  if (l == r)
    sum[v] = val;
  else {
    int m = (l + r) / 2;
    if (pos <= m)
      update(2 * v, l, m, pos, val);
    else
      update(2 * v + 1, m + 1, r, pos, val);
    sum[v] = sum[2 * v] + sum[2 * v + 1];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    int n;
    cin >> n;
    long long b[n];
    long long a[n + 1];
    a[0] = 0;
    for (int i = 0; i < n; i++) {
      cin >> b[i];
      a[i + 1] = i + 1;
    }
    build(a, 1, 0, n);
    for (int i = n - 1; i >= 0; i--) {
      long long l = 0;
      long long r = n;
      long long pos = -1;
      while (l <= r) {
        long long m = (l + r) / 2;
        long long val = query(1, 0, n, 0, m);
        if (val <= b[i]) {
          pos = m;
          l = m + 1;
        } else
          r = m - 1;
      }
      b[i] = pos + 1;
      update(1, 0, n, pos + 1, 0);
    }
    for (int i = 0; i < n; i++) {
      cout << b[i] << " ";
    }
    cout << "\n";
  }
  return 0;
}

```