#include <bits/stdc++.h>
using namespace std;
const long long inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long mod = 1e9 + 7;
long long tree[200050 << 2], lazy[200050 << 2], a[200050];
void pushup(long long rt) { tree[rt] = min(tree[rt << 1], tree[rt << 1 | 1]); }
void pushdown(long long rt) {
  if (lazy[rt]) {
    lazy[rt << 1] += lazy[rt];
    lazy[rt << 1 | 1] += lazy[rt];
    tree[rt << 1] -= lazy[rt];
    tree[rt << 1 | 1] -= lazy[rt];
    lazy[rt] = 0;
  }
}
void build(long long l, long long r, long long rt) {
  if (l == r) {
    tree[rt] = a[l];
    return;
  }
  long long mid = (l + r) / 2;
  build(l, mid, rt * 2);
  build(mid + 1, r, rt * 2 + 1);
  pushup(rt);
}
void update_id(long long pos, long long l, long long r, long long rt) {
  if (l == r) {
    tree[rt] = INF;
    return;
  }
  pushdown(rt);
  long long mid = (l + r) / 2;
  if (pos <= mid)
    update_id(pos, l, mid, rt * 2);
  else
    update_id(pos, mid + 1, r, rt * 2 + 1);
  pushup(rt);
}
void update_range(long long L, long long R, long long w, long long l,
                  long long r, long long rt) {
  if (L <= l && r <= R) {
    tree[rt] -= w;
    lazy[rt] += w;
    return;
  }
  pushdown(rt);
  long long mid = (l + r) / 2;
  if (L <= mid) update_range(L, R, w, l, mid, rt * 2);
  if (R > mid) update_range(L, R, w, mid + 1, r, rt * 2 + 1);
  pushup(rt);
}
long long query(long long l, long long r, long long rt) {
  if (l == r) {
    return l;
  }
  pushdown(rt);
  long long mid = (l + r) / 2;
  if (tree[rt << 1 | 1] == 0)
    return query(mid + 1, r, rt * 2 + 1);
  else
    return query(l, mid, rt * 2);
}
long long ans[200050];
int main() {
  long long n;
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  build(1, n, 1);
  for (long long i = 1; i <= n; i++) {
    int pos = query(1, n, 1);
    ans[pos] = i;
    update_id(pos, 1, n, 1);
    update_range(pos, n, i, 1, n, 1);
  }
  for (long long i = 1; i <= n; i++) cout << ans[i] << ' ';
  cout << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for 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 long long inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long mod = 1e9 + 7;
long long tree[200050 << 2], lazy[200050 << 2], a[200050];
void pushup(long long rt) { tree[rt] = min(tree[rt << 1], tree[rt << 1 | 1]); }
void pushdown(long long rt) {
  if (lazy[rt]) {
    lazy[rt << 1] += lazy[rt];
    lazy[rt << 1 | 1] += lazy[rt];
    tree[rt << 1] -= lazy[rt];
    tree[rt << 1 | 1] -= lazy[rt];
    lazy[rt] = 0;
  }
}
void build(long long l, long long r, long long rt) {
  if (l == r) {
    tree[rt] = a[l];
    return;
  }
  long long mid = (l + r) / 2;
  build(l, mid, rt * 2);
  build(mid + 1, r, rt * 2 + 1);
  pushup(rt);
}
void update_id(long long pos, long long l, long long r, long long rt) {
  if (l == r) {
    tree[rt] = INF;
    return;
  }
  pushdown(rt);
  long long mid = (l + r) / 2;
  if (pos <= mid)
    update_id(pos, l, mid, rt * 2);
  else
    update_id(pos, mid + 1, r, rt * 2 + 1);
  pushup(rt);
}
void update_range(long long L, long long R, long long w, long long l,
                  long long r, long long rt) {
  if (L <= l && r <= R) {
    tree[rt] -= w;
    lazy[rt] += w;
    return;
  }
  pushdown(rt);
  long long mid = (l + r) / 2;
  if (L <= mid) update_range(L, R, w, l, mid, rt * 2);
  if (R > mid) update_range(L, R, w, mid + 1, r, rt * 2 + 1);
  pushup(rt);
}
long long query(long long l, long long r, long long rt) {
  if (l == r) {
    return l;
  }
  pushdown(rt);
  long long mid = (l + r) / 2;
  if (tree[rt << 1 | 1] == 0)
    return query(mid + 1, r, rt * 2 + 1);
  else
    return query(l, mid, rt * 2);
}
long long ans[200050];
int main() {
  long long n;
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  build(1, n, 1);
  for (long long i = 1; i <= n; i++) {
    int pos = query(1, n, 1);
    ans[pos] = i;
    update_id(pos, 1, n, 1);
    update_range(pos, n, i, 1, n, 1);
  }
  for (long long i = 1; i <= n; i++) cout << ans[i] << ' ';
  cout << endl;
  return 0;
}

```