#include <bits/stdc++.h>
using namespace std;
struct S {
  int a, b;
  S() {}
  S(int _a, int _b) {
    a = _a;
    b = _b;
  }
  const bool operator<(const S &o) const { return a < o.a; }
};
string exm;
inline void exf(void) {
  cout << exm << "\n";
  exit(0);
}
template <typename T>
inline void showAll(vector<T> &v, string sep = "") {
  for (T &here : v) cout << here << sep;
}
template <typename T>
inline void showAll(T arr[], int st, int end, string sep = "") {
  for (int i = st; i <= end; i++) cout << arr[i] << sep;
}
template <typename T>
inline vector<int> int_seperation(T N, int d = 10) {
  vector<int> v;
  while (N) {
    v.push_back(N % d);
    N /= d;
  }
  reverse(v.begin(), v.end());
  return v;
}
const int SIZE = 200009;
long long arr[SIZE], tree[SIZE * 8];
int ans[SIZE];
int n;
long long getSum(int i) {
  long long res = 0;
  while (i) {
    res += tree[i];
    i -= (i & -i);
  }
  return res;
}
void update(int i, long long a) {
  while (i < n + 1) {
    tree[i] += a;
    i += (i & -i);
  }
  return;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &arr[i]);
  }
  for (int i = 1; i <= n; i++) {
    update(i, i);
  }
  for (int i = n; i; i--) {
    int l = 0, r = n + 1;
    while (l + 1 < r) {
      int mid = (l + r) / 2;
      long long sum = getSum(mid) - getSum(0);
      if (sum > arr[i])
        r = mid;
      else
        l = mid;
    }
    ans[i] = l + 1;
    update(l + 1, -l - 1);
  }
  showAll(ans, 1, n, " ");
  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>
using namespace std;
struct S {
  int a, b;
  S() {}
  S(int _a, int _b) {
    a = _a;
    b = _b;
  }
  const bool operator<(const S &o) const { return a < o.a; }
};
string exm;
inline void exf(void) {
  cout << exm << "\n";
  exit(0);
}
template <typename T>
inline void showAll(vector<T> &v, string sep = "") {
  for (T &here : v) cout << here << sep;
}
template <typename T>
inline void showAll(T arr[], int st, int end, string sep = "") {
  for (int i = st; i <= end; i++) cout << arr[i] << sep;
}
template <typename T>
inline vector<int> int_seperation(T N, int d = 10) {
  vector<int> v;
  while (N) {
    v.push_back(N % d);
    N /= d;
  }
  reverse(v.begin(), v.end());
  return v;
}
const int SIZE = 200009;
long long arr[SIZE], tree[SIZE * 8];
int ans[SIZE];
int n;
long long getSum(int i) {
  long long res = 0;
  while (i) {
    res += tree[i];
    i -= (i & -i);
  }
  return res;
}
void update(int i, long long a) {
  while (i < n + 1) {
    tree[i] += a;
    i += (i & -i);
  }
  return;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &arr[i]);
  }
  for (int i = 1; i <= n; i++) {
    update(i, i);
  }
  for (int i = n; i; i--) {
    int l = 0, r = n + 1;
    while (l + 1 < r) {
      int mid = (l + r) / 2;
      long long sum = getSum(mid) - getSum(0);
      if (sum > arr[i])
        r = mid;
      else
        l = mid;
    }
    ans[i] = l + 1;
    update(l + 1, -l - 1);
  }
  showAll(ans, 1, n, " ");
  return 0;
}

```