#include <bits/stdc++.h>
using namespace std;
void ECHO(string _s) {
  cout << endl;
  (void)_s;
}
template <typename T, typename... Args>
void ECHO(string _s, T x, Args... args) {
  int _i;
  string _s2 = "";
  for (_i = 0; _i < (int)(_s).size(); ++_i) {
    if (_s[_i] == ',') break;
    if (_s[_i] != ' ') _s2 += _s[_i];
  }
  if (_i == (int)(_s).size()) {
    --_i;
  }
  cout << "  (" << _s2 << "):" << x;
  ECHO(_s.substr(_i + 1, (int)(_s).size() - _i - 1), args...);
}
template <typename T0, typename T1>
inline ostream& operator<<(ostream& os, pair<T0, T1>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
inline ostream& operator<<(ostream& os, vector<T>& v) {
  for (int i = 0; i < (int)(v).size(); ++i) {
    if (i > 0) cout << "_";
    cout << v[i];
  }
  cout << endl;
  return os;
}
inline long long _gcd(long long a, long long b) {
  while (b) b %= a ^= b ^= a ^= b;
  return a;
}
mt19937 ran(time(0));
const long double PI = acos(-1);
const long double eps = 1e-9;
const long long inf = LLONG_MAX / 10;
vector<long long> v, ans;
long long n;
const long long MAX = 2 * 1e5 + 10;
vector<long long> bit(MAX, 0);
void update(int dx, long long val) {
  for (int i = dx + 1; i <= n; i += ((i) & (-(i)))) bit[i] += val;
}
long long query(int dx) {
  long long ans = 0, i;
  for (i = dx + 1; i > 0; i -= ((i) & (-(i)))) ans += bit[i];
  return ans;
}
int get_number(long long sum) {
  long long l = 0, r = n, mid, ans;
  while (l + 1 < r) {
    mid = (l + r) >> 1;
    ans = query(mid);
    if (ans <= sum)
      l = mid;
    else
      r = mid;
  }
  return l + 1;
}
int main() {
  ios::sync_with_stdio(false);
  int i, x;
  cin >> n;
  v.resize(n + 1);
  n = v.size() - 1;
  ans.resize(n + 1);
  for (i = 1; i <= n; ++i) {
    update(i, i);
    cin >> v[i];
  }
  for (i = n; i > 0; --i) {
    x = get_number(v[i]);
    ans[i] = x;
    update(x, -x);
  }
  for (i = 1; i <= n; ++i) {
    if (i > 1) cout << " ";
    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;
void ECHO(string _s) {
  cout << endl;
  (void)_s;
}
template <typename T, typename... Args>
void ECHO(string _s, T x, Args... args) {
  int _i;
  string _s2 = "";
  for (_i = 0; _i < (int)(_s).size(); ++_i) {
    if (_s[_i] == ',') break;
    if (_s[_i] != ' ') _s2 += _s[_i];
  }
  if (_i == (int)(_s).size()) {
    --_i;
  }
  cout << "  (" << _s2 << "):" << x;
  ECHO(_s.substr(_i + 1, (int)(_s).size() - _i - 1), args...);
}
template <typename T0, typename T1>
inline ostream& operator<<(ostream& os, pair<T0, T1>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
inline ostream& operator<<(ostream& os, vector<T>& v) {
  for (int i = 0; i < (int)(v).size(); ++i) {
    if (i > 0) cout << "_";
    cout << v[i];
  }
  cout << endl;
  return os;
}
inline long long _gcd(long long a, long long b) {
  while (b) b %= a ^= b ^= a ^= b;
  return a;
}
mt19937 ran(time(0));
const long double PI = acos(-1);
const long double eps = 1e-9;
const long long inf = LLONG_MAX / 10;
vector<long long> v, ans;
long long n;
const long long MAX = 2 * 1e5 + 10;
vector<long long> bit(MAX, 0);
void update(int dx, long long val) {
  for (int i = dx + 1; i <= n; i += ((i) & (-(i)))) bit[i] += val;
}
long long query(int dx) {
  long long ans = 0, i;
  for (i = dx + 1; i > 0; i -= ((i) & (-(i)))) ans += bit[i];
  return ans;
}
int get_number(long long sum) {
  long long l = 0, r = n, mid, ans;
  while (l + 1 < r) {
    mid = (l + r) >> 1;
    ans = query(mid);
    if (ans <= sum)
      l = mid;
    else
      r = mid;
  }
  return l + 1;
}
int main() {
  ios::sync_with_stdio(false);
  int i, x;
  cin >> n;
  v.resize(n + 1);
  n = v.size() - 1;
  ans.resize(n + 1);
  for (i = 1; i <= n; ++i) {
    update(i, i);
    cin >> v[i];
  }
  for (i = n; i > 0; --i) {
    x = get_number(v[i]);
    ans[i] = x;
    update(x, -x);
  }
  for (i = 1; i <= n; ++i) {
    if (i > 1) cout << " ";
    cout << ans[i];
  }
  cout << "\n";
  return 0;
}

```