#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f, N = 2e5 + 5;
inline int read() {
  int sum = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' && c <= '9') sum = sum * 10 + c - 48, c = getchar();
  return sum * f;
}
int n, a[N];
long long pre[N], c[N], s[N], ans[N];
bool vis[N];
void add(int x, int v) {
  for (int i = x; i <= n; i += (i & -i)) c[i] += v;
}
long long query(int x) {
  long long ans = 0;
  for (int i = x; i > 0; i -= (i & -i)) ans += c[i];
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> s[i];
  for (int i = 1; i <= n; i++) pre[i] = pre[i - 1] + i - 1;
  for (int i = n; i >= 1; i--) {
    int l = 1, r = n, pos = 0;
    while (l <= r) {
      int mid = (l + r) >> 1;
      long long tmp = query(mid);
      if (pre[mid] - tmp > s[i])
        r = mid - 1;
      else if (pre[mid] - tmp == s[i] && vis[mid])
        l = mid + 1;
      else if (pre[mid] - tmp == s[i]) {
        pos = mid;
        break;
      } else
        l = mid + 1;
    }
    ans[i] = pos;
    vis[pos] = 1;
    add(pos + 1, pos);
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  cout << '\n';
  return 0;
}

### Prompt

Generate 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 INF = 0x3f3f3f3f, N = 2e5 + 5;
inline int read() {
  int sum = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' && c <= '9') sum = sum * 10 + c - 48, c = getchar();
  return sum * f;
}
int n, a[N];
long long pre[N], c[N], s[N], ans[N];
bool vis[N];
void add(int x, int v) {
  for (int i = x; i <= n; i += (i & -i)) c[i] += v;
}
long long query(int x) {
  long long ans = 0;
  for (int i = x; i > 0; i -= (i & -i)) ans += c[i];
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> s[i];
  for (int i = 1; i <= n; i++) pre[i] = pre[i - 1] + i - 1;
  for (int i = n; i >= 1; i--) {
    int l = 1, r = n, pos = 0;
    while (l <= r) {
      int mid = (l + r) >> 1;
      long long tmp = query(mid);
      if (pre[mid] - tmp > s[i])
        r = mid - 1;
      else if (pre[mid] - tmp == s[i] && vis[mid])
        l = mid + 1;
      else if (pre[mid] - tmp == s[i]) {
        pos = mid;
        break;
      } else
        l = mid + 1;
    }
    ans[i] = pos;
    vis[pos] = 1;
    add(pos + 1, pos);
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  cout << '\n';
  return 0;
}

```