#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
int ans[maxn];
int n;
long long a[maxn];
const int maxnode = maxn << 2;
long long sum[maxnode], add[maxnode];
struct Node {
  int l, r;
  int mid() { return (l + r) >> 1; }
} node[maxnode];
void Pushup(int rt) { sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; }
void Pushdown(int rt, int m) {
  if (add[rt]) {
    add[rt << 1] += add[rt];
    add[rt << 1 | 1] += add[rt];
    sum[rt << 1] += add[rt] * (m - (m >> 1));
    sum[rt << 1 | 1] += add[rt] * (m >> 1);
    add[rt] = 0;
  }
}
void build(int l, int r, int rt) {
  node[rt].l = l;
  node[rt].r = r;
  add[rt] = 0;
  if (l == r) {
    sum[rt] = (long long)l;
    return;
  }
  int m = node[rt].mid();
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  Pushup(rt);
}
void update(int c, int l, int r, int rt) {
  if (node[rt].l == l && node[rt].r == r) {
    add[rt] += c;
    sum[rt] += (long long)c * (r - l + 1);
    return;
  }
  if (node[rt].l == node[rt].r) return;
  Pushdown(rt, node[rt].r - node[rt].l + 1);
  int m = node[rt].mid();
  if (r <= m)
    update(c, l, r, rt << 1);
  else if (l > m)
    update(c, l, r, rt << 1 | 1);
  else {
    update(c, l, m, rt << 1);
    update(c, m + 1, r, rt << 1 | 1);
  }
  Pushup(rt);
}
int res;
void query(int l, int r, int rt, long long x) {
  if (l == r) {
    res = l;
    return;
  }
  int m = node[rt].mid();
  Pushdown(rt, m);
  if (sum[rt << 1] > x && sum[rt << 1] > 0)
    query(l, m, rt << 1, x);
  else
    query(m + 1, r, rt << 1 | 1, x - sum[rt << 1]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  build(1, n, 1);
  for (int i = n; i; i--) {
    query(1, n, 1, a[i]);
    ans[i] = res;
    update(-res, res, res, 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 maxn = 2e5 + 10;
int ans[maxn];
int n;
long long a[maxn];
const int maxnode = maxn << 2;
long long sum[maxnode], add[maxnode];
struct Node {
  int l, r;
  int mid() { return (l + r) >> 1; }
} node[maxnode];
void Pushup(int rt) { sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; }
void Pushdown(int rt, int m) {
  if (add[rt]) {
    add[rt << 1] += add[rt];
    add[rt << 1 | 1] += add[rt];
    sum[rt << 1] += add[rt] * (m - (m >> 1));
    sum[rt << 1 | 1] += add[rt] * (m >> 1);
    add[rt] = 0;
  }
}
void build(int l, int r, int rt) {
  node[rt].l = l;
  node[rt].r = r;
  add[rt] = 0;
  if (l == r) {
    sum[rt] = (long long)l;
    return;
  }
  int m = node[rt].mid();
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  Pushup(rt);
}
void update(int c, int l, int r, int rt) {
  if (node[rt].l == l && node[rt].r == r) {
    add[rt] += c;
    sum[rt] += (long long)c * (r - l + 1);
    return;
  }
  if (node[rt].l == node[rt].r) return;
  Pushdown(rt, node[rt].r - node[rt].l + 1);
  int m = node[rt].mid();
  if (r <= m)
    update(c, l, r, rt << 1);
  else if (l > m)
    update(c, l, r, rt << 1 | 1);
  else {
    update(c, l, m, rt << 1);
    update(c, m + 1, r, rt << 1 | 1);
  }
  Pushup(rt);
}
int res;
void query(int l, int r, int rt, long long x) {
  if (l == r) {
    res = l;
    return;
  }
  int m = node[rt].mid();
  Pushdown(rt, m);
  if (sum[rt << 1] > x && sum[rt << 1] > 0)
    query(l, m, rt << 1, x);
  else
    query(m + 1, r, rt << 1 | 1, x - sum[rt << 1]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  build(1, n, 1);
  for (int i = n; i; i--) {
    query(1, n, 1, a[i]);
    ans[i] = res;
    update(-res, res, res, 1);
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}

```