#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
const long long inf = 1e18;
long long a[N], ans[N];
struct node {
  int l, r;
  long long lazy;
  long long v;
};
node e[N * 5];
void build(int root, int l, int r) {
  e[root].l = l;
  e[root].r = r;
  e[root].lazy = e[root].v = 0;
  if (l == r) {
    e[root].v = a[l];
    return;
  }
  build(root * 2, l, (l + r) / 2);
  build(root * 2 + 1, (l + r) / 2 + 1, r);
  e[root].v = min(e[root * 2].v, e[root * 2 + 1].v);
}
void updown(int root) {
  if (e[root].lazy) {
    e[root * 2].lazy += e[root].lazy;
    e[root * 2 + 1].lazy += e[root].lazy;
    e[root * 2].v -= e[root].lazy;
    e[root * 2 + 1].v -= e[root].lazy;
    e[root].lazy = 0;
  }
}
int que(int root, int x) {
  if (e[root].l == e[root].r) return e[root].l;
  updown(root);
  if (e[root * 2 + 1].v == x)
    return que(root * 2 + 1, x);
  else
    return que(root * 2, x);
}
void update1(int root, int x) {
  if (e[root].l == e[root].r) {
    e[root].v = inf;
    return;
  }
  int mid = (e[root].l + e[root].r) / 2;
  if (x <= mid)
    update1(root * 2, x);
  else
    update1(root * 2 + 1, x);
  e[root].v = min(e[root * 2].v, e[root * 2 + 1].v);
}
void update2(int root, int l, int r, int x) {
  if (e[root].l == l && e[root].r == r) {
    e[root].v -= x;
    e[root].lazy += x;
    return;
  }
  updown(root);
  int mid = (e[root].l + e[root].r) / 2;
  if (l > mid)
    update2(root * 2 + 1, l, r, x);
  else {
    if (r <= mid)
      update2(root * 2, l, r, x);
    else {
      update2(root * 2, l, mid, x);
      update2(root * 2 + 1, mid + 1, r, x);
    }
  }
  e[root].v = min(e[root * 2].v, e[root * 2 + 1].v);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    int k = que(1, 0);
    ans[k] = i;
    update1(1, k);
    if (k != n) update2(1, k + 1, n, i);
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}

### Prompt

Please formulate 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 N = 2e5 + 5;
const long long inf = 1e18;
long long a[N], ans[N];
struct node {
  int l, r;
  long long lazy;
  long long v;
};
node e[N * 5];
void build(int root, int l, int r) {
  e[root].l = l;
  e[root].r = r;
  e[root].lazy = e[root].v = 0;
  if (l == r) {
    e[root].v = a[l];
    return;
  }
  build(root * 2, l, (l + r) / 2);
  build(root * 2 + 1, (l + r) / 2 + 1, r);
  e[root].v = min(e[root * 2].v, e[root * 2 + 1].v);
}
void updown(int root) {
  if (e[root].lazy) {
    e[root * 2].lazy += e[root].lazy;
    e[root * 2 + 1].lazy += e[root].lazy;
    e[root * 2].v -= e[root].lazy;
    e[root * 2 + 1].v -= e[root].lazy;
    e[root].lazy = 0;
  }
}
int que(int root, int x) {
  if (e[root].l == e[root].r) return e[root].l;
  updown(root);
  if (e[root * 2 + 1].v == x)
    return que(root * 2 + 1, x);
  else
    return que(root * 2, x);
}
void update1(int root, int x) {
  if (e[root].l == e[root].r) {
    e[root].v = inf;
    return;
  }
  int mid = (e[root].l + e[root].r) / 2;
  if (x <= mid)
    update1(root * 2, x);
  else
    update1(root * 2 + 1, x);
  e[root].v = min(e[root * 2].v, e[root * 2 + 1].v);
}
void update2(int root, int l, int r, int x) {
  if (e[root].l == l && e[root].r == r) {
    e[root].v -= x;
    e[root].lazy += x;
    return;
  }
  updown(root);
  int mid = (e[root].l + e[root].r) / 2;
  if (l > mid)
    update2(root * 2 + 1, l, r, x);
  else {
    if (r <= mid)
      update2(root * 2, l, r, x);
    else {
      update2(root * 2, l, mid, x);
      update2(root * 2 + 1, mid + 1, r, x);
    }
  }
  e[root].v = min(e[root * 2].v, e[root * 2 + 1].v);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    int k = que(1, 0);
    ans[k] = i;
    update1(1, k);
    if (k != n) update2(1, k + 1, n, i);
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}

```