#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (isdigit(ch)) s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
inline void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const int maxn = 2e5 + 10;
long long a[maxn], coun[maxn << 2], sum[maxn << 2];
long long ans[maxn];
int n;
void build(int l, int r, int i) {
  if (l == r) {
    sum[i] = 1;
    coun[i] = l;
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, i << 1);
  build(mid + 1, r, (i << 1) + 1);
  sum[i] = sum[i << 1] + sum[(i << 1) + 1];
  coun[i] = coun[i << 1] + coun[(i << 1) + 1];
}
void update(int l, int r, int pos, int i) {
  if (l == r) {
    sum[i] = 0;
    coun[i] = 0;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid) update(l, mid, pos, i << 1);
  if (pos > mid) update(mid + 1, r, pos, (i << 1) + 1);
  sum[i] = sum[i << 1] + sum[(i << 1) + 1];
  coun[i] = coun[i << 1] + coun[(i << 1) + 1];
}
long long query(int l, int r, int i, int k) {
  if (sum[i] <= k) return coun[i];
  int mid = (l + r) >> 1;
  if (sum[i << 1] >= k) return query(l, mid, i << 1, k);
  return coun[i << 1] + query(mid + 1, r, (i << 1) + 1, k - sum[i << 1]);
}
int queryMin(int l, int r, int i, int k) {
  if (l == r) return l;
  int mid = (l + r) >> 1;
  if (sum[i << 1] >= k) return queryMin(l, mid, i << 1, k);
  return queryMin(mid + 1, r, (i << 1) + 1, k - sum[i << 1]);
}
bool check(int mid, long long x) { return query(1, n, 1, mid) >= x; }
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  build(1, n, 1);
  for (int i = n; i >= 1; --i) {
    int l = 1, r = n;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (check(mid - 1, a[i])) {
        ans[i] = queryMin(1, n, 1, mid);
        r = mid - 1;
      } else
        l = mid + 1;
    }
    update(1, n, ans[i], 1);
  }
  for (int i = 1; i <= n; ++i) write(ans[i]), putchar(' ');
  putchar('\n');
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
inline long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (isdigit(ch)) s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
inline void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const int maxn = 2e5 + 10;
long long a[maxn], coun[maxn << 2], sum[maxn << 2];
long long ans[maxn];
int n;
void build(int l, int r, int i) {
  if (l == r) {
    sum[i] = 1;
    coun[i] = l;
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, i << 1);
  build(mid + 1, r, (i << 1) + 1);
  sum[i] = sum[i << 1] + sum[(i << 1) + 1];
  coun[i] = coun[i << 1] + coun[(i << 1) + 1];
}
void update(int l, int r, int pos, int i) {
  if (l == r) {
    sum[i] = 0;
    coun[i] = 0;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid) update(l, mid, pos, i << 1);
  if (pos > mid) update(mid + 1, r, pos, (i << 1) + 1);
  sum[i] = sum[i << 1] + sum[(i << 1) + 1];
  coun[i] = coun[i << 1] + coun[(i << 1) + 1];
}
long long query(int l, int r, int i, int k) {
  if (sum[i] <= k) return coun[i];
  int mid = (l + r) >> 1;
  if (sum[i << 1] >= k) return query(l, mid, i << 1, k);
  return coun[i << 1] + query(mid + 1, r, (i << 1) + 1, k - sum[i << 1]);
}
int queryMin(int l, int r, int i, int k) {
  if (l == r) return l;
  int mid = (l + r) >> 1;
  if (sum[i << 1] >= k) return queryMin(l, mid, i << 1, k);
  return queryMin(mid + 1, r, (i << 1) + 1, k - sum[i << 1]);
}
bool check(int mid, long long x) { return query(1, n, 1, mid) >= x; }
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  build(1, n, 1);
  for (int i = n; i >= 1; --i) {
    int l = 1, r = n;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (check(mid - 1, a[i])) {
        ans[i] = queryMin(1, n, 1, mid);
        r = mid - 1;
      } else
        l = mid + 1;
    }
    update(1, n, ans[i], 1);
  }
  for (int i = 1; i <= n; ++i) write(ans[i]), putchar(' ');
  putchar('\n');
  return 0;
}

```