#include <bits/stdc++.h>
const long long inf = 0x3f3f3f3f;
const long long inn = 0x80808080;
using namespace std;
const long long maxm = 2e5 + 5;
long long mi[maxm << 2];
long long laz[maxm << 2];
long long p[maxm];
long long n;
void pushup(long long node) { mi[node] = min(mi[node * 2], mi[node * 2 + 1]); }
void pushdown(long long node) {
  if (laz[node]) {
    mi[node * 2] += laz[node];
    mi[node * 2 + 1] += laz[node];
    laz[node * 2] += laz[node];
    laz[node * 2 + 1] += laz[node];
    laz[node] = 0;
  }
}
void build(long long l, long long r, long long node) {
  if (l == r) {
    cin >> mi[node];
    return;
  }
  long long mid = (l + r) / 2;
  build(l, mid, node * 2);
  build(mid + 1, r, node * 2 + 1);
  pushup(node);
}
long long ask(long long x, long long l, long long r, long long node) {
  if (l == r) {
    mi[node] = (1LL << 60);
    return l;
  }
  long long mid = (l + r) / 2;
  pushdown(node);
  long long ans = 1;
  if (mi[node * 2 + 1] <= x)
    ans = ask(x, mid + 1, r, node * 2 + 1);
  else
    ans = ask(x, l, mid, node * 2);
  pushup(node);
  return ans;
}
void update(long long st, long long ed, long long val, long long l, long long r,
            long long node) {
  if (st <= l && ed >= r) {
    mi[node] += val;
    laz[node] += val;
    return;
  }
  pushdown(node);
  long long mid = (l + r) / 2;
  if (st <= mid) update(st, ed, val, l, mid, node * 2);
  if (ed >= mid) update(st, ed, val, mid + 1, r, node * 2 + 1);
  pushup(node);
}
signed main() {
  ios::sync_with_stdio(0);
  cin >> n;
  build(1, n, 1);
  for (long long i = 1; i <= n; i++) {
    long long x = ask(0, 1, n, 1);
    p[x] = i;
    if (x != n) {
      update(x + 1, n, -i, 1, n, 1);
    }
  }
  for (long long i = 1; i <= n; i++) {
    cout << p[i] << ' ';
  }
  cout << endl;
  return 0;
}

### 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>
const long long inf = 0x3f3f3f3f;
const long long inn = 0x80808080;
using namespace std;
const long long maxm = 2e5 + 5;
long long mi[maxm << 2];
long long laz[maxm << 2];
long long p[maxm];
long long n;
void pushup(long long node) { mi[node] = min(mi[node * 2], mi[node * 2 + 1]); }
void pushdown(long long node) {
  if (laz[node]) {
    mi[node * 2] += laz[node];
    mi[node * 2 + 1] += laz[node];
    laz[node * 2] += laz[node];
    laz[node * 2 + 1] += laz[node];
    laz[node] = 0;
  }
}
void build(long long l, long long r, long long node) {
  if (l == r) {
    cin >> mi[node];
    return;
  }
  long long mid = (l + r) / 2;
  build(l, mid, node * 2);
  build(mid + 1, r, node * 2 + 1);
  pushup(node);
}
long long ask(long long x, long long l, long long r, long long node) {
  if (l == r) {
    mi[node] = (1LL << 60);
    return l;
  }
  long long mid = (l + r) / 2;
  pushdown(node);
  long long ans = 1;
  if (mi[node * 2 + 1] <= x)
    ans = ask(x, mid + 1, r, node * 2 + 1);
  else
    ans = ask(x, l, mid, node * 2);
  pushup(node);
  return ans;
}
void update(long long st, long long ed, long long val, long long l, long long r,
            long long node) {
  if (st <= l && ed >= r) {
    mi[node] += val;
    laz[node] += val;
    return;
  }
  pushdown(node);
  long long mid = (l + r) / 2;
  if (st <= mid) update(st, ed, val, l, mid, node * 2);
  if (ed >= mid) update(st, ed, val, mid + 1, r, node * 2 + 1);
  pushup(node);
}
signed main() {
  ios::sync_with_stdio(0);
  cin >> n;
  build(1, n, 1);
  for (long long i = 1; i <= n; i++) {
    long long x = ask(0, 1, n, 1);
    p[x] = i;
    if (x != n) {
      update(x + 1, n, -i, 1, n, 1);
    }
  }
  for (long long i = 1; i <= n; i++) {
    cout << p[i] << ' ';
  }
  cout << endl;
  return 0;
}

```