#include <bits/stdc++.h>
using namespace std;
const int maxn = 200005;
const long long inf = 1e11;
struct Node {
  int l, r, num;
  long long tag;
  long long val;
} node[4 * maxn];
long long a[maxn];
int ans[maxn];
void up(int no) {
  if (node[no << 1].val < node[no << 1 | 1].val) {
    node[no].val = node[no << 1].val;
    node[no].num = node[no << 1].num;
  } else {
    node[no].val = node[no << 1 | 1].val;
    node[no].num = node[no << 1 | 1].num;
  }
}
void build(int l, int r, int no) {
  node[no].l = l;
  node[no].r = r;
  node[no].tag = 0;
  if (l == r) {
    node[no].val = a[l];
    node[no].num = l;
    return;
  }
  int mid = l + r >> 1;
  build(l, mid, no << 1);
  build(mid + 1, r, no << 1 | 1);
  up(no);
}
void pushdown(int no) {
  node[no << 1].tag += node[no].tag;
  node[no << 1].val += node[no].tag;
  node[no << 1 | 1].tag += node[no].tag;
  node[no << 1 | 1].val += node[no].tag;
  node[no].tag = 0;
}
void change(int l, int r, int no, long long num) {
  if (node[no].r < l || node[no].l > r) return;
  if (node[no].l >= l && node[no].r <= r) {
    node[no].tag += num;
    node[no].val += num;
    return;
  }
  if (node[no].tag != 0) {
    pushdown(no);
  }
  change(l, r, no << 1, num);
  change(l, r, no << 1 | 1, num);
  up(no);
}
int main() {
  int n, i;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%lld", &a[i]);
  build(1, n, 1);
  for (i = 1; i <= n; i++) {
    int x = node[1].num;
    ans[x] = i;
    change(x, x, 1, inf);
    change(x + 1, n, 1, -i);
  }
  for (i = 1; i <= n; i++) {
    printf("%d ", ans[i]);
  }
  printf("\n");
}

### Prompt

Please create a solution in cpp 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 maxn = 200005;
const long long inf = 1e11;
struct Node {
  int l, r, num;
  long long tag;
  long long val;
} node[4 * maxn];
long long a[maxn];
int ans[maxn];
void up(int no) {
  if (node[no << 1].val < node[no << 1 | 1].val) {
    node[no].val = node[no << 1].val;
    node[no].num = node[no << 1].num;
  } else {
    node[no].val = node[no << 1 | 1].val;
    node[no].num = node[no << 1 | 1].num;
  }
}
void build(int l, int r, int no) {
  node[no].l = l;
  node[no].r = r;
  node[no].tag = 0;
  if (l == r) {
    node[no].val = a[l];
    node[no].num = l;
    return;
  }
  int mid = l + r >> 1;
  build(l, mid, no << 1);
  build(mid + 1, r, no << 1 | 1);
  up(no);
}
void pushdown(int no) {
  node[no << 1].tag += node[no].tag;
  node[no << 1].val += node[no].tag;
  node[no << 1 | 1].tag += node[no].tag;
  node[no << 1 | 1].val += node[no].tag;
  node[no].tag = 0;
}
void change(int l, int r, int no, long long num) {
  if (node[no].r < l || node[no].l > r) return;
  if (node[no].l >= l && node[no].r <= r) {
    node[no].tag += num;
    node[no].val += num;
    return;
  }
  if (node[no].tag != 0) {
    pushdown(no);
  }
  change(l, r, no << 1, num);
  change(l, r, no << 1 | 1, num);
  up(no);
}
int main() {
  int n, i;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%lld", &a[i]);
  build(1, n, 1);
  for (i = 1; i <= n; i++) {
    int x = node[1].num;
    ans[x] = i;
    change(x, x, 1, inf);
    change(x + 1, n, 1, -i);
  }
  for (i = 1; i <= n; i++) {
    printf("%d ", ans[i]);
  }
  printf("\n");
}

```