#include <bits/stdc++.h>
using namespace std;
long int n;
long long int tree[200001 * 10];
long int a[200002];
void build(long int node, long int start, long int end) {
  if (start == end) {
    tree[node] = a[start];
  } else {
    long int mid = (start + end) / 2;
    build(2 * node + 1, start, mid);
    build(2 * node + 2, mid + 1, end);
    tree[node] = tree[2 * node + 1] + tree[2 * node + 2];
  }
}
void update(long int node, long int start, long int end, long int ind) {
  if (start == end)
    tree[node] = a[ind];
  else {
    long int mid = (start + end) / 2;
    if (start <= ind && ind <= mid) {
      update(2 * node + 1, start, mid, ind);
    } else {
      update(2 * node + 2, mid + 1, end, ind);
    }
    tree[node] = tree[2 * node + 1] + tree[2 * node + 2];
  }
}
long long int query(long int node, long int start, long int end, long int l,
                    long int r) {
  if (r < start || end < l) {
    return 0;
  }
  if (l <= start && end <= r) {
    return tree[node];
  }
  long int mid = (start + end) / 2;
  long long int p1 = query(2 * node + 1, start, mid, l, r);
  long long int p2 = query(2 * node + 2, mid + 1, end, l, r);
  return (p1 + p2);
}
long int idx[200002];
long int root(long int x) {
  while (idx[x] != x) {
    idx[x] = idx[idx[x]];
    x = idx[x];
  }
  return x;
}
int main() {
  cin >> n;
  long int ans = 0;
  for (long int i = 0; i <= n + 1; i++) a[i] = i, idx[i] = i;
  build(0, 0, n);
  long long int b[n];
  for (long int i = 0; i < n; i++) cin >> b[i];
  vector<long int> x;
  for (long int i = n - 1; i >= 0; i--) {
    long int l = 0, r = n;
    long int ans = -1;
    while (l <= r) {
      long int mid = (l + r) / 2;
      long long int temp1 = query(0, 0, n, 0, mid);
      if (temp1 == b[i]) {
        ans = mid + 1;
        break;
      } else if (temp1 < b[i]) {
        l = mid + 1;
      } else
        r = mid - 1;
    }
    ans = root(ans);
    x.push_back(ans);
    a[ans] = 0;
    update(0, 0, n, ans);
    idx[root(ans)] = root(ans + 1);
  }
  for (long int i = 0; i < x.size(); i++) cout << x[x.size() - 1 - i] << " ";
}

### Prompt

Your task is to create 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;
long int n;
long long int tree[200001 * 10];
long int a[200002];
void build(long int node, long int start, long int end) {
  if (start == end) {
    tree[node] = a[start];
  } else {
    long int mid = (start + end) / 2;
    build(2 * node + 1, start, mid);
    build(2 * node + 2, mid + 1, end);
    tree[node] = tree[2 * node + 1] + tree[2 * node + 2];
  }
}
void update(long int node, long int start, long int end, long int ind) {
  if (start == end)
    tree[node] = a[ind];
  else {
    long int mid = (start + end) / 2;
    if (start <= ind && ind <= mid) {
      update(2 * node + 1, start, mid, ind);
    } else {
      update(2 * node + 2, mid + 1, end, ind);
    }
    tree[node] = tree[2 * node + 1] + tree[2 * node + 2];
  }
}
long long int query(long int node, long int start, long int end, long int l,
                    long int r) {
  if (r < start || end < l) {
    return 0;
  }
  if (l <= start && end <= r) {
    return tree[node];
  }
  long int mid = (start + end) / 2;
  long long int p1 = query(2 * node + 1, start, mid, l, r);
  long long int p2 = query(2 * node + 2, mid + 1, end, l, r);
  return (p1 + p2);
}
long int idx[200002];
long int root(long int x) {
  while (idx[x] != x) {
    idx[x] = idx[idx[x]];
    x = idx[x];
  }
  return x;
}
int main() {
  cin >> n;
  long int ans = 0;
  for (long int i = 0; i <= n + 1; i++) a[i] = i, idx[i] = i;
  build(0, 0, n);
  long long int b[n];
  for (long int i = 0; i < n; i++) cin >> b[i];
  vector<long int> x;
  for (long int i = n - 1; i >= 0; i--) {
    long int l = 0, r = n;
    long int ans = -1;
    while (l <= r) {
      long int mid = (l + r) / 2;
      long long int temp1 = query(0, 0, n, 0, mid);
      if (temp1 == b[i]) {
        ans = mid + 1;
        break;
      } else if (temp1 < b[i]) {
        l = mid + 1;
      } else
        r = mid - 1;
    }
    ans = root(ans);
    x.push_back(ans);
    a[ans] = 0;
    update(0, 0, n, ans);
    idx[root(ans)] = root(ans + 1);
  }
  for (long int i = 0; i < x.size(); i++) cout << x[x.size() - 1 - i] << " ";
}

```