#include <bits/stdc++.h>
using namespace std;
template <class T>
bool tomin(T &x, T y) {
  return y < x ? x = y, 1 : 0;
}
template <class T>
bool tomax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
template <class T>
void read(T &x) {
  char c;
  x = 0;
  int f = 1;
  while (c = getchar(), c < '0' || c > '9')
    if (c == '-') f = -1;
  do x = (x << 3) + (x << 1) + (c ^ 48);
  while (c = getchar(), c >= '0' && c <= '9');
  x *= f;
}
bool mem1;
const double Pi = acos(-1);
const int maxn = 2e5 + 5;
int n;
long long A[maxn];
int ans[maxn];
long long c[maxn];
void Add(int x, int v) {
  while (x <= n) {
    c[x] += v;
    x += x & -x;
  }
}
long long query(int x) {
  long long res = 0;
  while (x) {
    res += c[x];
    x &= x - 1;
  }
  return res;
}
bool mem2;
int main() {
  srand(time(NULL));
  read(n);
  for (int i = 1, i_ = n; i <= i_; ++i) read(A[i]);
  for (int i = 1, i_ = n; i <= i_; ++i) Add(i, i);
  for (int i = n, i_ = 1; i >= i_; --i) {
    int L = 0, R = n - 1, mid, res;
    while (L <= R) {
      if (query(mid = L + R >> 1) <= A[i])
        L = (res = mid) + 1;
      else
        R = mid - 1;
    }
    ans[i] = res + 1;
    Add(ans[i], -ans[i]);
  }
  for (int i = 1, i_ = n; i <= i_; ++i) printf("%d ", ans[i]);
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
template <class T>
bool tomin(T &x, T y) {
  return y < x ? x = y, 1 : 0;
}
template <class T>
bool tomax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
template <class T>
void read(T &x) {
  char c;
  x = 0;
  int f = 1;
  while (c = getchar(), c < '0' || c > '9')
    if (c == '-') f = -1;
  do x = (x << 3) + (x << 1) + (c ^ 48);
  while (c = getchar(), c >= '0' && c <= '9');
  x *= f;
}
bool mem1;
const double Pi = acos(-1);
const int maxn = 2e5 + 5;
int n;
long long A[maxn];
int ans[maxn];
long long c[maxn];
void Add(int x, int v) {
  while (x <= n) {
    c[x] += v;
    x += x & -x;
  }
}
long long query(int x) {
  long long res = 0;
  while (x) {
    res += c[x];
    x &= x - 1;
  }
  return res;
}
bool mem2;
int main() {
  srand(time(NULL));
  read(n);
  for (int i = 1, i_ = n; i <= i_; ++i) read(A[i]);
  for (int i = 1, i_ = n; i <= i_; ++i) Add(i, i);
  for (int i = n, i_ = 1; i >= i_; --i) {
    int L = 0, R = n - 1, mid, res;
    while (L <= R) {
      if (query(mid = L + R >> 1) <= A[i])
        L = (res = mid) + 1;
      else
        R = mid - 1;
    }
    ans[i] = res + 1;
    Add(ans[i], -ans[i]);
  }
  for (int i = 1, i_ = n; i <= i_; ++i) printf("%d ", ans[i]);
  return 0;
}

```