#include <bits/stdc++.h>
using namespace std;
int const maxn = 1e6 + 5;
long long tree[4 * maxn], lazy[4 * maxn], a[maxn];
int kq[maxn];
int n;
void down(int g) {
  if (lazy[g] != 0) {
    tree[g << 1] += lazy[g];
    lazy[g << 1] += lazy[g];
    tree[(g << 1) | 1] += lazy[g];
    lazy[(g << 1) | 1] += lazy[g];
    lazy[g] = 0;
  }
}
void buld(int g, int l, int r) {
  if (l == r) {
    tree[g] = a[l];
  } else {
    int m = (l + r) >> 1;
    buld(g << 1, l, m);
    buld((g << 1) | 1, m + 1, r);
    tree[g] = min(tree[g << 1], tree[(g << 1) | 1]);
  }
}
void update(int g, int l, int r, int u, int v, long long val) {
  if (r < u || v < l) return;
  if (u <= l && r <= v) {
    tree[g] += val;
    lazy[g] += val;
  } else {
    down(g);
    int m = (l + r) >> 1;
    update(g << 1, l, m, u, v, val);
    update((g << 1) | 1, m + 1, r, u, v, val);
    tree[g] = min(tree[g << 1], tree[(g << 1) | 1]);
  }
}
int find_zero(int g, int l, int r) {
  if (l == r)
    return l;
  else {
    down(g);
    int m = (l + r) >> 1;
    int res = -1;
    if (tree[(g << 1) | 1] == 0)
      res = find_zero((g << 1) | 1, m + 1, r);
    else
      res = find_zero(g << 1, l, m);
    tree[g] = min(tree[g << 1], tree[(g << 1) | 1]);
    return res;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  buld(1, 1, n);
  for (int i = 1; i <= n; i++) {
    int pos = find_zero(1, 1, n);
    kq[pos] = i;
    update(1, 1, n, pos, pos, 1e18);
    update(1, 1, n, pos + 1, n, -i);
  }
  for (int i = 1; i <= n; i++) cout << kq[i] << " ";
  return 0;
}

### Prompt

Your challenge is to write 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;
int const maxn = 1e6 + 5;
long long tree[4 * maxn], lazy[4 * maxn], a[maxn];
int kq[maxn];
int n;
void down(int g) {
  if (lazy[g] != 0) {
    tree[g << 1] += lazy[g];
    lazy[g << 1] += lazy[g];
    tree[(g << 1) | 1] += lazy[g];
    lazy[(g << 1) | 1] += lazy[g];
    lazy[g] = 0;
  }
}
void buld(int g, int l, int r) {
  if (l == r) {
    tree[g] = a[l];
  } else {
    int m = (l + r) >> 1;
    buld(g << 1, l, m);
    buld((g << 1) | 1, m + 1, r);
    tree[g] = min(tree[g << 1], tree[(g << 1) | 1]);
  }
}
void update(int g, int l, int r, int u, int v, long long val) {
  if (r < u || v < l) return;
  if (u <= l && r <= v) {
    tree[g] += val;
    lazy[g] += val;
  } else {
    down(g);
    int m = (l + r) >> 1;
    update(g << 1, l, m, u, v, val);
    update((g << 1) | 1, m + 1, r, u, v, val);
    tree[g] = min(tree[g << 1], tree[(g << 1) | 1]);
  }
}
int find_zero(int g, int l, int r) {
  if (l == r)
    return l;
  else {
    down(g);
    int m = (l + r) >> 1;
    int res = -1;
    if (tree[(g << 1) | 1] == 0)
      res = find_zero((g << 1) | 1, m + 1, r);
    else
      res = find_zero(g << 1, l, m);
    tree[g] = min(tree[g << 1], tree[(g << 1) | 1]);
    return res;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  buld(1, 1, n);
  for (int i = 1; i <= n; i++) {
    int pos = find_zero(1, 1, n);
    kq[pos] = i;
    update(1, 1, n, pos, pos, 1e18);
    update(1, 1, n, pos + 1, n, -i);
  }
  for (int i = 1; i <= n; i++) cout << kq[i] << " ";
  return 0;
}

```