#include <bits/stdc++.h>
using namespace std;
struct node {
  int64_t data, lazy;
  int pos;
};
node st[200001 * 4];
int64_t input[200001];
int anw[200001];
void output(int root, int L, int R) {
  if (L == R) {
    return;
  }
  int mid = (L + R) >> 1;
  output(root << 1, L, mid);
  output(root << 1 | 1, mid + 1, R);
}
void Init(int root, int L, int R) {
  st[root].lazy = 0;
  if (L == R) {
    st[root].data = input[L];
    st[root].pos = L;
    return;
  }
  int mid = (L + R) >> 1;
  Init(root << 1, L, mid);
  Init(root << 1 | 1, mid + 1, R);
  int lpos = root << 1, rpos = root << 1 | 1;
  st[root].data = min(st[lpos].data, st[rpos].data);
  if (st[root].data == st[rpos].data) {
    st[root].pos = st[rpos].pos;
  } else {
    st[root].pos = st[lpos].pos;
  }
}
void Update(int root, int l, int r, int L, int R, int64_t val) {
  if (l == L && r == R) {
    st[root].lazy += val;
    st[root].data += val;
    return;
  }
  int mid = (L + R) >> 1;
  if (st[root].lazy != 0) {
    Update(root << 1, L, mid, L, mid, st[root].lazy);
    Update(root << 1 | 1, mid + 1, R, mid + 1, R, st[root].lazy);
    st[root].lazy = 0;
  }
  if (r <= mid) {
    Update(root << 1, l, r, L, mid, val);
  } else if (mid < l) {
    Update(root << 1 | 1, l, r, mid + 1, R, val);
  } else {
    Update(root << 1, l, mid, L, mid, val);
    Update(root << 1 | 1, mid + 1, r, mid + 1, R, val);
  }
  int lpos = root << 1, rpos = root << 1 | 1;
  st[root].data = min(st[lpos].data, st[rpos].data);
  if (st[root].data == st[rpos].data) {
    st[root].pos = st[rpos].pos;
  } else {
    st[root].pos = st[lpos].pos;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> input[i];
  }
  Init(1, 1, n);
  int64_t MAX = int64_t(n) * int64_t(n) * int64_t(100);
  if (n == 2) {
    output(1, 1, n);
  }
  for (int i = 1; i <= n; i++) {
    int pos = st[1].pos;
    anw[pos] = i;
    Update(1, pos, pos, 1, n, MAX);
    Update(1, pos, n, 1, n, int64_t(i * -1));
    if (n == 2) {
      output(1, 1, n);
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << anw[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;
struct node {
  int64_t data, lazy;
  int pos;
};
node st[200001 * 4];
int64_t input[200001];
int anw[200001];
void output(int root, int L, int R) {
  if (L == R) {
    return;
  }
  int mid = (L + R) >> 1;
  output(root << 1, L, mid);
  output(root << 1 | 1, mid + 1, R);
}
void Init(int root, int L, int R) {
  st[root].lazy = 0;
  if (L == R) {
    st[root].data = input[L];
    st[root].pos = L;
    return;
  }
  int mid = (L + R) >> 1;
  Init(root << 1, L, mid);
  Init(root << 1 | 1, mid + 1, R);
  int lpos = root << 1, rpos = root << 1 | 1;
  st[root].data = min(st[lpos].data, st[rpos].data);
  if (st[root].data == st[rpos].data) {
    st[root].pos = st[rpos].pos;
  } else {
    st[root].pos = st[lpos].pos;
  }
}
void Update(int root, int l, int r, int L, int R, int64_t val) {
  if (l == L && r == R) {
    st[root].lazy += val;
    st[root].data += val;
    return;
  }
  int mid = (L + R) >> 1;
  if (st[root].lazy != 0) {
    Update(root << 1, L, mid, L, mid, st[root].lazy);
    Update(root << 1 | 1, mid + 1, R, mid + 1, R, st[root].lazy);
    st[root].lazy = 0;
  }
  if (r <= mid) {
    Update(root << 1, l, r, L, mid, val);
  } else if (mid < l) {
    Update(root << 1 | 1, l, r, mid + 1, R, val);
  } else {
    Update(root << 1, l, mid, L, mid, val);
    Update(root << 1 | 1, mid + 1, r, mid + 1, R, val);
  }
  int lpos = root << 1, rpos = root << 1 | 1;
  st[root].data = min(st[lpos].data, st[rpos].data);
  if (st[root].data == st[rpos].data) {
    st[root].pos = st[rpos].pos;
  } else {
    st[root].pos = st[lpos].pos;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> input[i];
  }
  Init(1, 1, n);
  int64_t MAX = int64_t(n) * int64_t(n) * int64_t(100);
  if (n == 2) {
    output(1, 1, n);
  }
  for (int i = 1; i <= n; i++) {
    int pos = st[1].pos;
    anw[pos] = i;
    Update(1, pos, pos, 1, n, MAX);
    Update(1, pos, n, 1, n, int64_t(i * -1));
    if (n == 2) {
      output(1, 1, n);
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << anw[i] << " ";
  }
  return 0;
}

```