#include <bits/stdc++.h>
using namespace std;
const int M = 2e5 + 5;
int p[M << 2], res[M];
long long V[M << 2], tag[M << 2];
void push_up(int o) {
  if (V[o << 1] < V[o << 1 | 1]) {
    V[o] = V[o << 1];
    p[o] = p[o << 1];
  } else {
    V[o] = V[o << 1 | 1];
    p[o] = p[o << 1 | 1];
  }
}
void push_down(int o) {
  tag[o << 1] += tag[o];
  tag[o << 1 | 1] += tag[o];
  V[o << 1] += tag[o];
  V[o << 1 | 1] += tag[o];
  tag[o] = 0;
}
void build(int o, int l, int r) {
  if (l == r) {
    scanf("%lld", &V[o]);
    p[o] = l;
    return;
  }
  int mid = (l + r) / 2;
  build(o << 1, l, mid);
  build(o << 1 | 1, mid + 1, r);
  push_up(o);
}
void upd(int o, int l, int r, int L, int R, long long x) {
  if (L > R) return;
  if (L <= l && r <= R) {
    tag[o] += x;
    V[o] += x;
    return;
  }
  int mid = (l + r) / 2;
  push_down(o);
  if (L <= mid) upd(o << 1, l, mid, L, R, x);
  if (R > mid) upd(o << 1 | 1, mid + 1, r, L, R, x);
  push_up(o);
}
int main() {
  int n;
  scanf("%d", &n);
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    int pos = p[1];
    res[pos] = i;
    upd(1, 1, n, pos, pos, 1e11);
    upd(1, 1, n, pos + 1, n, -i);
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", res[i]);
  }
  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;
const int M = 2e5 + 5;
int p[M << 2], res[M];
long long V[M << 2], tag[M << 2];
void push_up(int o) {
  if (V[o << 1] < V[o << 1 | 1]) {
    V[o] = V[o << 1];
    p[o] = p[o << 1];
  } else {
    V[o] = V[o << 1 | 1];
    p[o] = p[o << 1 | 1];
  }
}
void push_down(int o) {
  tag[o << 1] += tag[o];
  tag[o << 1 | 1] += tag[o];
  V[o << 1] += tag[o];
  V[o << 1 | 1] += tag[o];
  tag[o] = 0;
}
void build(int o, int l, int r) {
  if (l == r) {
    scanf("%lld", &V[o]);
    p[o] = l;
    return;
  }
  int mid = (l + r) / 2;
  build(o << 1, l, mid);
  build(o << 1 | 1, mid + 1, r);
  push_up(o);
}
void upd(int o, int l, int r, int L, int R, long long x) {
  if (L > R) return;
  if (L <= l && r <= R) {
    tag[o] += x;
    V[o] += x;
    return;
  }
  int mid = (l + r) / 2;
  push_down(o);
  if (L <= mid) upd(o << 1, l, mid, L, R, x);
  if (R > mid) upd(o << 1 | 1, mid + 1, r, L, R, x);
  push_up(o);
}
int main() {
  int n;
  scanf("%d", &n);
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    int pos = p[1];
    res[pos] = i;
    upd(1, 1, n, pos, pos, 1e11);
    upd(1, 1, n, pos + 1, n, -i);
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", res[i]);
  }
  return 0;
}

```