#include <bits/stdc++.h>
using namespace std;
long long n, s[200005], tree[800005], lazy[800005], ans[200005];
void pushdown(long long ind, long long l, long long r) {
  if (l != r && lazy[ind] != 0) {
    lazy[2 * ind] += lazy[ind];
    tree[2 * ind] += lazy[ind];
    lazy[2 * ind + 1] += lazy[ind];
    tree[2 * ind + 1] += lazy[ind];
    lazy[ind] = 0;
  }
}
void build(long long ind, long long l, long long r) {
  if (l == r) {
    tree[ind] = s[l];
    return;
  }
  long long mid = (l + r) / 2;
  build(2 * ind, l, mid);
  build(2 * ind + 1, mid + 1, r);
  tree[ind] = min(tree[2 * ind], tree[2 * ind + 1]);
}
long long query(long long ind, long long l, long long r) {
  pushdown(ind, l, r);
  if (l == r) return l;
  tree[ind] = min(tree[2 * ind], tree[2 * ind + 1]);
  long long mid = (l + r) / 2;
  if (tree[2 * ind + 1] == 0)
    query(2 * ind + 1, mid + 1, r);
  else
    query(2 * ind, l, mid);
}
void update(long long ind, long long l, long long r, long long ll, long long rr,
            long long val) {
  pushdown(ind, l, r);
  if (r < ll || l > rr) return;
  if (ll <= l && r <= rr) {
    tree[ind] += val;
    lazy[ind] += val;
    return;
  }
  long long mid = (l + r) / 2;
  update(2 * ind, l, mid, ll, rr, val);
  update(2 * ind + 1, mid + 1, r, ll, rr, val);
  tree[ind] = min(tree[2 * ind], tree[2 * ind + 1]);
}
int32_t main() {
  cin >> n;
  for (long long i = 1; i <= n; ++i) cin >> s[i];
  build(1, 1, n);
  for (long long i = 1; i <= n; ++i) {
    long long t = query(1, 1, n);
    ans[t] = i;
    if (t < n) update(1, 1, n, t + 1, n, -i);
    update(1, 1, n, t, t, 1e18);
  }
  for (long long i = 1; i <= n; ++i) cout << ans[i] << " ";
  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;
long long n, s[200005], tree[800005], lazy[800005], ans[200005];
void pushdown(long long ind, long long l, long long r) {
  if (l != r && lazy[ind] != 0) {
    lazy[2 * ind] += lazy[ind];
    tree[2 * ind] += lazy[ind];
    lazy[2 * ind + 1] += lazy[ind];
    tree[2 * ind + 1] += lazy[ind];
    lazy[ind] = 0;
  }
}
void build(long long ind, long long l, long long r) {
  if (l == r) {
    tree[ind] = s[l];
    return;
  }
  long long mid = (l + r) / 2;
  build(2 * ind, l, mid);
  build(2 * ind + 1, mid + 1, r);
  tree[ind] = min(tree[2 * ind], tree[2 * ind + 1]);
}
long long query(long long ind, long long l, long long r) {
  pushdown(ind, l, r);
  if (l == r) return l;
  tree[ind] = min(tree[2 * ind], tree[2 * ind + 1]);
  long long mid = (l + r) / 2;
  if (tree[2 * ind + 1] == 0)
    query(2 * ind + 1, mid + 1, r);
  else
    query(2 * ind, l, mid);
}
void update(long long ind, long long l, long long r, long long ll, long long rr,
            long long val) {
  pushdown(ind, l, r);
  if (r < ll || l > rr) return;
  if (ll <= l && r <= rr) {
    tree[ind] += val;
    lazy[ind] += val;
    return;
  }
  long long mid = (l + r) / 2;
  update(2 * ind, l, mid, ll, rr, val);
  update(2 * ind + 1, mid + 1, r, ll, rr, val);
  tree[ind] = min(tree[2 * ind], tree[2 * ind + 1]);
}
int32_t main() {
  cin >> n;
  for (long long i = 1; i <= n; ++i) cin >> s[i];
  build(1, 1, n);
  for (long long i = 1; i <= n; ++i) {
    long long t = query(1, 1, n);
    ans[t] = i;
    if (t < n) update(1, 1, n, t + 1, n, -i);
    update(1, 1, n, t, t, 1e18);
  }
  for (long long i = 1; i <= n; ++i) cout << ans[i] << " ";
  return 0;
}

```