#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b, long long m) {
  a %= m;
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ((ret % m) * (a % m)) % m;
    a = ((a % m) * (a % m)) % m;
    b >>= 1;
  }
  return ret;
}
const int N = 2e5 + 5;
long long n, bit[N + 3], a[N + 2], ans[N + 2];
void update(long long val, int x) {
  for (; x <= N; x += x & (-x)) {
    bit[x] += val;
  }
}
long long query(int x) {
  int tm = x;
  long long ret = 0;
  while (x > 0) {
    ret += bit[x];
    x -= x & (-x);
  }
  ret += ans[tm - 1];
  return ret;
}
void pre() {
  long long now = 0;
  ans[0] = 0;
  for (long long i = 1; i < N; i++) {
    now += i;
    ans[i] = now;
  }
}
int main() {
  pre();
  cin >> n;
  vector<long long> vec;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = n - 1; i >= 0; i--) {
    int l = 1, r = n + 1, mid;
    while ((r - l) > 1) {
      mid = (l + r) / 2;
      long long result = query(mid);
      if (result <= a[i]) {
        l = mid;
      } else {
        r = mid;
      }
    }
    if (query(l + 1) == a[i]) {
      vec.push_back(l + 1);
      update(-(l + 1), l + 2);
    } else {
      vec.push_back(l);
      update(-(l), l + 1);
    }
  }
  reverse(vec.begin(), vec.end());
  for (auto it : vec) {
    cout << it << " ";
  }
  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;
long long power(long long a, long long b, long long m) {
  a %= m;
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ((ret % m) * (a % m)) % m;
    a = ((a % m) * (a % m)) % m;
    b >>= 1;
  }
  return ret;
}
const int N = 2e5 + 5;
long long n, bit[N + 3], a[N + 2], ans[N + 2];
void update(long long val, int x) {
  for (; x <= N; x += x & (-x)) {
    bit[x] += val;
  }
}
long long query(int x) {
  int tm = x;
  long long ret = 0;
  while (x > 0) {
    ret += bit[x];
    x -= x & (-x);
  }
  ret += ans[tm - 1];
  return ret;
}
void pre() {
  long long now = 0;
  ans[0] = 0;
  for (long long i = 1; i < N; i++) {
    now += i;
    ans[i] = now;
  }
}
int main() {
  pre();
  cin >> n;
  vector<long long> vec;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = n - 1; i >= 0; i--) {
    int l = 1, r = n + 1, mid;
    while ((r - l) > 1) {
      mid = (l + r) / 2;
      long long result = query(mid);
      if (result <= a[i]) {
        l = mid;
      } else {
        r = mid;
      }
    }
    if (query(l + 1) == a[i]) {
      vec.push_back(l + 1);
      update(-(l + 1), l + 2);
    } else {
      vec.push_back(l);
      update(-(l), l + 1);
    }
  }
  reverse(vec.begin(), vec.end());
  for (auto it : vec) {
    cout << it << " ";
  }
  return 0;
}

```