#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ '0');
    ch = getchar();
  }
  return x * f;
}
namespace BIT {
long long C[1000005];
inline void Add(long long x, long long val) {
  for (register long long i = x; i < 1000005; i += (i & (-i))) {
    C[i] += val;
  }
}
inline long long Query(long long x) {
  long long ans = 0;
  for (register long long i = x; i > 0; i -= (i & (-i))) {
    ans += C[i];
  }
  return ans;
}
};  // namespace BIT
using namespace BIT;
long long a[1000005], n;
inline long long BinSearch(long long pos) {
  long long l = 0, r = 1000005 - 1, ans;
  while (l < r - 1) {
    long long mid = (l + r) >> 1;
    if (Query(mid) <= a[pos])
      l = mid;
    else
      r = mid;
  }
  return r;
}
long long ans[1000005];
int main() {
  n = read();
  for (register long long i = 1; i <= n; ++i) a[i] = read();
  for (register long long i = 1; i <= n; ++i) {
    Add(i, i);
  }
  for (register long long i = n; i >= 1; --i) {
    long long pos = BinSearch(i);
    Add(pos, -pos);
    ans[i] = pos;
  }
  for (register long long i = 1; i <= n; ++i) {
    printf("%I64d ", ans[i]);
  }
}

### Prompt

In CPP, your task is to solve 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;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ '0');
    ch = getchar();
  }
  return x * f;
}
namespace BIT {
long long C[1000005];
inline void Add(long long x, long long val) {
  for (register long long i = x; i < 1000005; i += (i & (-i))) {
    C[i] += val;
  }
}
inline long long Query(long long x) {
  long long ans = 0;
  for (register long long i = x; i > 0; i -= (i & (-i))) {
    ans += C[i];
  }
  return ans;
}
};  // namespace BIT
using namespace BIT;
long long a[1000005], n;
inline long long BinSearch(long long pos) {
  long long l = 0, r = 1000005 - 1, ans;
  while (l < r - 1) {
    long long mid = (l + r) >> 1;
    if (Query(mid) <= a[pos])
      l = mid;
    else
      r = mid;
  }
  return r;
}
long long ans[1000005];
int main() {
  n = read();
  for (register long long i = 1; i <= n; ++i) a[i] = read();
  for (register long long i = 1; i <= n; ++i) {
    Add(i, i);
  }
  for (register long long i = n; i >= 1; --i) {
    long long pos = BinSearch(i);
    Add(pos, -pos);
    ans[i] = pos;
  }
  for (register long long i = 1; i <= n; ++i) {
    printf("%I64d ", ans[i]);
  }
}

```