#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
const long long INF = 1e18;
int n;
long long a[maxn], ans[maxn];
long long tree[maxn << 2], laz[maxn << 2];
void pushup(int rt) { tree[rt] = min(tree[rt << 1], tree[(rt << 1) | 1]); }
void build(int l, int r, int rt) {
  if (l == r) {
    tree[rt] = a[l];
    return;
  }
  build(l, ((l + r) >> 1), 2 * rt);
  build(((l + r) >> 1) + 1, r, 2 * rt + 1);
  pushup(rt);
}
void change(int x, int l, int r, int rt) {
  if (l == r) {
    tree[rt] = INF;
    return;
  }
  if (x <= ((l + r) >> 1))
    change(x, l, ((l + r) >> 1), 2 * rt);
  else
    change(x, ((l + r) >> 1) + 1, r, 2 * rt + 1);
  pushup(rt);
}
void pushdown(int rt) {
  long long& x = laz[rt];
  if (x) {
    tree[rt << 1] += x;
    tree[(rt << 1) | 1] += x;
    laz[rt << 1] += x;
    laz[(rt << 1) | 1] += x;
    x = 0;
  }
}
void update(int x, int ql, int qr, int l, int r, int rt) {
  if (ql == l && qr == r) {
    laz[rt] += x;
    tree[rt] += x;
    return;
  }
  pushdown(rt);
  if (qr <= ((l + r) >> 1))
    update(x, ql, qr, l, ((l + r) >> 1), 2 * rt);
  else if (ql > ((l + r) >> 1))
    update(x, ql, qr, ((l + r) >> 1) + 1, r, 2 * rt + 1);
  else {
    update(x, ql, ((l + r) >> 1), l, ((l + r) >> 1), 2 * rt);
    update(x, ((l + r) >> 1) + 1, qr, ((l + r) >> 1) + 1, r, 2 * rt + 1);
  }
  pushup(rt);
}
int query(int l, int r, int rt) {
  if (l == r) return l;
  pushdown(rt);
  if (tree[((l + r) >> 1) + 1, r, 2 * rt + 1] == 0)
    return query(((l + r) >> 1) + 1, r, 2 * rt + 1);
  else
    return query(l, ((l + r) >> 1), 2 * rt);
}
int main() {
  cin.sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  build(1, n, 1);
  int t;
  for (int i = 1; i <= n; i++) {
    t = query(1, n, 1);
    ans[t] = i;
    change(t, 1, n, 1);
    if (t != n) update(-i, t + 1, n, 1, n, 1);
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  return 0;
}

### Prompt

Develop a solution in Cpp 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 + 5;
const long long INF = 1e18;
int n;
long long a[maxn], ans[maxn];
long long tree[maxn << 2], laz[maxn << 2];
void pushup(int rt) { tree[rt] = min(tree[rt << 1], tree[(rt << 1) | 1]); }
void build(int l, int r, int rt) {
  if (l == r) {
    tree[rt] = a[l];
    return;
  }
  build(l, ((l + r) >> 1), 2 * rt);
  build(((l + r) >> 1) + 1, r, 2 * rt + 1);
  pushup(rt);
}
void change(int x, int l, int r, int rt) {
  if (l == r) {
    tree[rt] = INF;
    return;
  }
  if (x <= ((l + r) >> 1))
    change(x, l, ((l + r) >> 1), 2 * rt);
  else
    change(x, ((l + r) >> 1) + 1, r, 2 * rt + 1);
  pushup(rt);
}
void pushdown(int rt) {
  long long& x = laz[rt];
  if (x) {
    tree[rt << 1] += x;
    tree[(rt << 1) | 1] += x;
    laz[rt << 1] += x;
    laz[(rt << 1) | 1] += x;
    x = 0;
  }
}
void update(int x, int ql, int qr, int l, int r, int rt) {
  if (ql == l && qr == r) {
    laz[rt] += x;
    tree[rt] += x;
    return;
  }
  pushdown(rt);
  if (qr <= ((l + r) >> 1))
    update(x, ql, qr, l, ((l + r) >> 1), 2 * rt);
  else if (ql > ((l + r) >> 1))
    update(x, ql, qr, ((l + r) >> 1) + 1, r, 2 * rt + 1);
  else {
    update(x, ql, ((l + r) >> 1), l, ((l + r) >> 1), 2 * rt);
    update(x, ((l + r) >> 1) + 1, qr, ((l + r) >> 1) + 1, r, 2 * rt + 1);
  }
  pushup(rt);
}
int query(int l, int r, int rt) {
  if (l == r) return l;
  pushdown(rt);
  if (tree[((l + r) >> 1) + 1, r, 2 * rt + 1] == 0)
    return query(((l + r) >> 1) + 1, r, 2 * rt + 1);
  else
    return query(l, ((l + r) >> 1), 2 * rt);
}
int main() {
  cin.sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  build(1, n, 1);
  int t;
  for (int i = 1; i <= n; i++) {
    t = query(1, n, 1);
    ans[t] = i;
    change(t, 1, n, 1);
    if (t != n) update(-i, t + 1, n, 1, n, 1);
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  return 0;
}

```