#include <bits/stdc++.h>
using namespace std;
const int M = 2e5 + 10;
const long long inf = 1e15 + 10;
const long long mod = 1e9 + 7;
set<int> s;
long long a[M], ans[M];
long long sum[4 * M], lazy[4 * M];
int v[M * 4];
void pushup(int x) { sum[x] = min(sum[x << 1], sum[x << 1 | 1]); }
void built(int l, int r, int i) {
  if (l == r) {
    sum[i] = a[l];
    return;
  }
  int mid = l + r >> 1;
  built(l, mid, i << 1);
  built(mid + 1, r, i << 1 | 1);
  pushup(i);
}
void pp(int i) {
  if (lazy[i]) {
    sum[i << 1] -= lazy[i];
    sum[i << 1 | 1] -= lazy[i];
    lazy[i << 1] += lazy[i];
    lazy[i << 1 | 1] += lazy[i];
    lazy[i] = 0;
  }
}
void add(int l, int r, int x, int y, long long va, int i) {
  if (l >= x && r <= y) {
    sum[i] -= va;
    lazy[i] += va;
    return;
  }
  pp(i);
  int mid = l + r >> 1;
  if (x <= mid) add(l, mid, x, y, va, i << 1);
  if (y > mid) add(mid + 1, r, x, y, va, i << 1 | 1);
  pushup(i);
}
long long query(int l, int r, int i) {
  if (l == r) {
    return l;
  }
  pp(i);
  int mid = l + r >> 1, k;
  if (sum[i << 1] < sum[i << 1 | 1])
    k = query(l, mid, i << 1);
  else
    k = query(mid + 1, r, i << 1 | 1);
  return k;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  built(1, n, 1);
  for (int i = 1; i <= n; i++) {
    int pos = query(1, n, 1);
    ans[pos] = i;
    add(1, n, pos, pos, -inf, 1);
    add(1, n, pos + 1, n, i, 1);
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", ans[i]);
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
const int M = 2e5 + 10;
const long long inf = 1e15 + 10;
const long long mod = 1e9 + 7;
set<int> s;
long long a[M], ans[M];
long long sum[4 * M], lazy[4 * M];
int v[M * 4];
void pushup(int x) { sum[x] = min(sum[x << 1], sum[x << 1 | 1]); }
void built(int l, int r, int i) {
  if (l == r) {
    sum[i] = a[l];
    return;
  }
  int mid = l + r >> 1;
  built(l, mid, i << 1);
  built(mid + 1, r, i << 1 | 1);
  pushup(i);
}
void pp(int i) {
  if (lazy[i]) {
    sum[i << 1] -= lazy[i];
    sum[i << 1 | 1] -= lazy[i];
    lazy[i << 1] += lazy[i];
    lazy[i << 1 | 1] += lazy[i];
    lazy[i] = 0;
  }
}
void add(int l, int r, int x, int y, long long va, int i) {
  if (l >= x && r <= y) {
    sum[i] -= va;
    lazy[i] += va;
    return;
  }
  pp(i);
  int mid = l + r >> 1;
  if (x <= mid) add(l, mid, x, y, va, i << 1);
  if (y > mid) add(mid + 1, r, x, y, va, i << 1 | 1);
  pushup(i);
}
long long query(int l, int r, int i) {
  if (l == r) {
    return l;
  }
  pp(i);
  int mid = l + r >> 1, k;
  if (sum[i << 1] < sum[i << 1 | 1])
    k = query(l, mid, i << 1);
  else
    k = query(mid + 1, r, i << 1 | 1);
  return k;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  built(1, n, 1);
  for (int i = 1; i <= n; i++) {
    int pos = query(1, n, 1);
    ans[pos] = i;
    add(1, n, pos, pos, -inf, 1);
    add(1, n, pos + 1, n, i, 1);
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", ans[i]);
  }
  return 0;
}

```