#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
const int maxm = 4e6 + 10;
const long long mod = 1e9 + 7;
const int inf = 0x3ffffff;
const double eps = 1e-5;
long long n, s[maxn];
long long a[maxn], ans[maxn << 2], tag[maxn << 2], m;
long long res[maxn];
inline void push_up(long long p) { ans[p] = ans[p << 1] + ans[p << 1 | 1]; }
inline void f(long long p, long long l, long long r, long long k) {
  tag[p] += k;
  ans[p] += (r - l + 1) * k;
}
void build(long long l, long long r, long long p) {
  if (l == r) {
    ans[p] = a[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build(l, mid, p << 1);
  build(mid + 1, r, p << 1 | 1);
  push_up(p);
}
inline void push_down(long long p, long long l, long long r) {
  long long mid = (l + r) >> 1;
  f(p << 1, l, mid, tag[p]);
  f(p << 1 | 1, mid + 1, r, tag[p]);
  tag[p] = 0;
}
inline void update(long long nl, long long nr, long long l, long long r,
                   long long p, long long k) {
  if (nl <= l && r <= nr) {
    ans[p] += k * (r - l + 1);
    tag[p] += k;
    return;
  }
  push_down(p, l, r);
  long long mid = (l + r) >> 1;
  if (nl <= mid) update(nl, nr, l, mid, p << 1, k);
  if (nr > mid) update(nl, nr, mid + 1, r, p << 1 | 1, k);
  push_up(p);
}
long long query(long long q_x, long long q_y, long long l, long long r,
                long long p) {
  long long res = 0;
  if (q_x <= l && q_y >= r) return ans[p];
  long long mid = (l + r) >> 1;
  push_down(p, l, r);
  if (q_x <= mid) res += query(q_x, q_y, l, mid, p << 1);
  if (q_y > mid) res += query(q_x, q_y, mid + 1, r, p << 1 | 1);
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    a[i] = i;
  }
  build(1, n, 1);
  for (int i = n; i >= 1; i--) {
    int l = 1, r = n;
    while (l < r) {
      int mid = (l + r) >> 1;
      if (query(1, mid, 1, n, 1) > s[i])
        r = mid;
      else
        l = mid + 1;
    }
    update(l, l, 1, n, 1, -l);
    res[i] = l;
  }
  for (int i = 1; i <= n; i++) cout << res[i] << " ";
  return ~~(0 - 0);
}

### Prompt

Please formulate 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;
const int maxn = 2e5 + 10;
const int maxm = 4e6 + 10;
const long long mod = 1e9 + 7;
const int inf = 0x3ffffff;
const double eps = 1e-5;
long long n, s[maxn];
long long a[maxn], ans[maxn << 2], tag[maxn << 2], m;
long long res[maxn];
inline void push_up(long long p) { ans[p] = ans[p << 1] + ans[p << 1 | 1]; }
inline void f(long long p, long long l, long long r, long long k) {
  tag[p] += k;
  ans[p] += (r - l + 1) * k;
}
void build(long long l, long long r, long long p) {
  if (l == r) {
    ans[p] = a[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build(l, mid, p << 1);
  build(mid + 1, r, p << 1 | 1);
  push_up(p);
}
inline void push_down(long long p, long long l, long long r) {
  long long mid = (l + r) >> 1;
  f(p << 1, l, mid, tag[p]);
  f(p << 1 | 1, mid + 1, r, tag[p]);
  tag[p] = 0;
}
inline void update(long long nl, long long nr, long long l, long long r,
                   long long p, long long k) {
  if (nl <= l && r <= nr) {
    ans[p] += k * (r - l + 1);
    tag[p] += k;
    return;
  }
  push_down(p, l, r);
  long long mid = (l + r) >> 1;
  if (nl <= mid) update(nl, nr, l, mid, p << 1, k);
  if (nr > mid) update(nl, nr, mid + 1, r, p << 1 | 1, k);
  push_up(p);
}
long long query(long long q_x, long long q_y, long long l, long long r,
                long long p) {
  long long res = 0;
  if (q_x <= l && q_y >= r) return ans[p];
  long long mid = (l + r) >> 1;
  push_down(p, l, r);
  if (q_x <= mid) res += query(q_x, q_y, l, mid, p << 1);
  if (q_y > mid) res += query(q_x, q_y, mid + 1, r, p << 1 | 1);
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    a[i] = i;
  }
  build(1, n, 1);
  for (int i = n; i >= 1; i--) {
    int l = 1, r = n;
    while (l < r) {
      int mid = (l + r) >> 1;
      if (query(1, mid, 1, n, 1) > s[i])
        r = mid;
      else
        l = mid + 1;
    }
    update(l, l, 1, n, 1, -l);
    res[i] = l;
  }
  for (int i = 1; i <= n; i++) cout << res[i] << " ";
  return ~~(0 - 0);
}

```