#include <bits/stdc++.h>
using namespace std;
long long A[300001];
long long tree[300001 * 4 + 2], prop[4 * 300001 + 2], ans[300001];
void create(long long node, long long b, long long e) {
  if (b == e) {
    tree[node] = A[b];
    return;
  }
  long long l = 2 * node;
  long long r = l + 1;
  long long m = (b + e) / 2;
  create(l, b, m);
  create(r, m + 1, e);
  tree[node] = min(tree[l], tree[r]);
}
void update(long long node, long long b, long long e, long long i, long long j,
            long long val) {
  if (b > j || e < i) return;
  if (i <= b && e <= j) {
    if (val == ((long long)1 << 62)) {
      tree[node] = ((long long)1 << 62);
      return;
    }
    prop[node] += val;
    tree[node] -= val;
    return;
  }
  long long l = 2 * node;
  long long r = l + 1;
  long long m = (b + e) / 2;
  if (prop[node]) {
    prop[l] += prop[node];
    prop[r] += prop[node];
    tree[l] = -prop[node];
    tree[r] -= prop[node];
    prop[node] = 0;
  }
  update(l, b, m, i, j, val);
  update(r, m + 1, e, i, j, val);
  tree[node] = min(tree[l], tree[r]);
}
long long query(long long node, long long b, long long e) {
  if (b == e) return b;
  long long l = 2 * node;
  long long r = l + 1;
  long long m = (b + e) / 2;
  if (prop[node]) {
    prop[l] += prop[node];
    prop[r] += prop[node];
    tree[l] -= prop[node];
    tree[r] -= prop[node];
    prop[node] = 0;
  }
  if (tree[r] == 0)
    return query(r, m + 1, e);
  else
    return query(l, b, m);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  memset(prop, 0, sizeof prop);
  ;
  long long n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> A[i];
  create(1, 1, n);
  for (int i = 1; i <= n; i++) {
    long long p = query(1, 1, n);
    ans[p] = i;
    update(1, 1, n, p, p, ((long long)1 << 62));
    update(1, 1, n, p + 1, n, i);
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << "\n";
}

### Prompt

Your challenge is to write 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;
long long A[300001];
long long tree[300001 * 4 + 2], prop[4 * 300001 + 2], ans[300001];
void create(long long node, long long b, long long e) {
  if (b == e) {
    tree[node] = A[b];
    return;
  }
  long long l = 2 * node;
  long long r = l + 1;
  long long m = (b + e) / 2;
  create(l, b, m);
  create(r, m + 1, e);
  tree[node] = min(tree[l], tree[r]);
}
void update(long long node, long long b, long long e, long long i, long long j,
            long long val) {
  if (b > j || e < i) return;
  if (i <= b && e <= j) {
    if (val == ((long long)1 << 62)) {
      tree[node] = ((long long)1 << 62);
      return;
    }
    prop[node] += val;
    tree[node] -= val;
    return;
  }
  long long l = 2 * node;
  long long r = l + 1;
  long long m = (b + e) / 2;
  if (prop[node]) {
    prop[l] += prop[node];
    prop[r] += prop[node];
    tree[l] = -prop[node];
    tree[r] -= prop[node];
    prop[node] = 0;
  }
  update(l, b, m, i, j, val);
  update(r, m + 1, e, i, j, val);
  tree[node] = min(tree[l], tree[r]);
}
long long query(long long node, long long b, long long e) {
  if (b == e) return b;
  long long l = 2 * node;
  long long r = l + 1;
  long long m = (b + e) / 2;
  if (prop[node]) {
    prop[l] += prop[node];
    prop[r] += prop[node];
    tree[l] -= prop[node];
    tree[r] -= prop[node];
    prop[node] = 0;
  }
  if (tree[r] == 0)
    return query(r, m + 1, e);
  else
    return query(l, b, m);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  memset(prop, 0, sizeof prop);
  ;
  long long n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> A[i];
  create(1, 1, n);
  for (int i = 1; i <= n; i++) {
    long long p = query(1, 1, n);
    ans[p] = i;
    update(1, 1, n, p, p, ((long long)1 << 62));
    update(1, 1, n, p + 1, n, i);
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << "\n";
}

```