#include <bits/stdc++.h>
#pragma GCC optimize "03"
using namespace std;
const long long int N = 2e5 + 5;
const long long int mod = 1e9 + 7;
const long long int inf = 1e18 + 9;
long long int a[N];
long long int t[4 * N], lzy[4 * N];
void build(long long int nd, long long int s, long long int e) {
  if (s == e)
    t[nd] = a[s], a[s] = 0;
  else {
    long long int md = (s + e) >> 1;
    build(2 * nd, s, md);
    build(2 * nd + 1, md + 1, e);
    t[nd] = min(t[2 * nd], t[2 * nd + 1]);
  }
}
void pushdown(long long int nd, long long int s, long long int e) {
  if (s == e) {
    t[nd] -= lzy[nd];
    lzy[nd] = 0;
    return;
  }
  lzy[2 * nd] += lzy[nd];
  lzy[2 * nd + 1] += lzy[nd];
  t[nd] -= lzy[nd];
  lzy[nd] = 0;
}
void upd(long long int nd, long long int s, long long int e, long long int l,
         long long int r, long long int v) {
  if (s > r || e < l || l > r) return;
  if (s >= l && e <= r) {
    lzy[nd] += v;
    return;
  }
  pushdown(nd, s, e);
  long long int md = (s + e) >> 1;
  upd(2 * nd, s, md, l, r, v);
  upd(2 * nd + 1, md + 1, e, l, r, v);
  t[nd] = min(t[2 * nd] - lzy[2 * nd], t[2 * nd + 1] - lzy[2 * nd + 1]);
}
long long int query(long long int nd, long long int s, long long int e) {
  pushdown(nd, s, e);
  if (s == e) return s;
  long long int md = (s + e) >> 1;
  long long int p;
  if (t[2 * nd + 1] - lzy[2 * nd + 1] == 0)
    p = query(2 * nd + 1, md + 1, e);
  else
    p = query(2 * nd, s, md);
  t[nd] = min(t[2 * nd] - lzy[2 * nd], t[2 * nd + 1] - lzy[2 * nd + 1]);
  return p;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  for (long long int i = 1; i <= n; i++) cin >> a[i];
  build(1, 1, n);
  for (long long int i = 1; i <= n; i++) {
    long long int id = query(1, 1, n);
    a[id] = i;
    upd(1, 1, n, id, id, -inf);
    upd(1, 1, n, id + 1, n, i);
  }
  for (long long int i = 1; i <= n; i++) cout << a[i] << " ";
  return 0;
}

### Prompt

Please create a solution in Cpp 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>
#pragma GCC optimize "03"
using namespace std;
const long long int N = 2e5 + 5;
const long long int mod = 1e9 + 7;
const long long int inf = 1e18 + 9;
long long int a[N];
long long int t[4 * N], lzy[4 * N];
void build(long long int nd, long long int s, long long int e) {
  if (s == e)
    t[nd] = a[s], a[s] = 0;
  else {
    long long int md = (s + e) >> 1;
    build(2 * nd, s, md);
    build(2 * nd + 1, md + 1, e);
    t[nd] = min(t[2 * nd], t[2 * nd + 1]);
  }
}
void pushdown(long long int nd, long long int s, long long int e) {
  if (s == e) {
    t[nd] -= lzy[nd];
    lzy[nd] = 0;
    return;
  }
  lzy[2 * nd] += lzy[nd];
  lzy[2 * nd + 1] += lzy[nd];
  t[nd] -= lzy[nd];
  lzy[nd] = 0;
}
void upd(long long int nd, long long int s, long long int e, long long int l,
         long long int r, long long int v) {
  if (s > r || e < l || l > r) return;
  if (s >= l && e <= r) {
    lzy[nd] += v;
    return;
  }
  pushdown(nd, s, e);
  long long int md = (s + e) >> 1;
  upd(2 * nd, s, md, l, r, v);
  upd(2 * nd + 1, md + 1, e, l, r, v);
  t[nd] = min(t[2 * nd] - lzy[2 * nd], t[2 * nd + 1] - lzy[2 * nd + 1]);
}
long long int query(long long int nd, long long int s, long long int e) {
  pushdown(nd, s, e);
  if (s == e) return s;
  long long int md = (s + e) >> 1;
  long long int p;
  if (t[2 * nd + 1] - lzy[2 * nd + 1] == 0)
    p = query(2 * nd + 1, md + 1, e);
  else
    p = query(2 * nd, s, md);
  t[nd] = min(t[2 * nd] - lzy[2 * nd], t[2 * nd + 1] - lzy[2 * nd + 1]);
  return p;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  for (long long int i = 1; i <= n; i++) cin >> a[i];
  build(1, 1, n);
  for (long long int i = 1; i <= n; i++) {
    long long int id = query(1, 1, n);
    a[id] = i;
    upd(1, 1, n, id, id, -inf);
    upd(1, 1, n, id + 1, n, i);
  }
  for (long long int i = 1; i <= n; i++) cout << a[i] << " ";
  return 0;
}

```