#include <bits/stdc++.h>
using namespace std;
using ll = long long;
inline ll read();
const int M = 500016, MOD = 1000000007;
struct BinaryIndexTree {
  ll bit[M], n;
  inline void modify(ll p, ll x) {
    for (; p <= n; p += p & -p) bit[p] += x;
  }
  inline ll sum(ll p) {
    ll res = 0;
    for (; p; p -= p & -p) res += bit[p];
    return res;
  }
  inline ll sum(ll l, ll r) { return sum(r) - sum(l - 1); }
} bit;
ll s[M], save[M], vis[M];
int cal(ll s, int n) {
  ll lef = 1, rig = n;
  while (lef <= rig) {
    ll mid = (lef + rig) >> 1;
    ll val = mid * (mid - 1) / 2 - bit.sum(mid - 1);
    if (val == s && !vis[mid])
      return mid;
    else if (val <= s)
      lef = mid + 1;
    else
      rig = mid - 1;
  }
  assert(0);
  return -1;
}
int main(void) {
  int n = read();
  bit.n = n;
  for (int i = 1; i <= n; ++i) s[i] = read();
  for (int i = n; i >= 1; --i) {
    save[i] = cal(s[i], n);
    bit.modify(save[i], save[i]);
    vis[save[i]] = 1;
  }
  for (int i = 1; i <= n; ++i) cout << save[i] << " ";
  cout << endl;
  return 0;
}
inline ll read() {
  ll x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}

### Prompt

Generate 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;
using ll = long long;
inline ll read();
const int M = 500016, MOD = 1000000007;
struct BinaryIndexTree {
  ll bit[M], n;
  inline void modify(ll p, ll x) {
    for (; p <= n; p += p & -p) bit[p] += x;
  }
  inline ll sum(ll p) {
    ll res = 0;
    for (; p; p -= p & -p) res += bit[p];
    return res;
  }
  inline ll sum(ll l, ll r) { return sum(r) - sum(l - 1); }
} bit;
ll s[M], save[M], vis[M];
int cal(ll s, int n) {
  ll lef = 1, rig = n;
  while (lef <= rig) {
    ll mid = (lef + rig) >> 1;
    ll val = mid * (mid - 1) / 2 - bit.sum(mid - 1);
    if (val == s && !vis[mid])
      return mid;
    else if (val <= s)
      lef = mid + 1;
    else
      rig = mid - 1;
  }
  assert(0);
  return -1;
}
int main(void) {
  int n = read();
  bit.n = n;
  for (int i = 1; i <= n; ++i) s[i] = read();
  for (int i = n; i >= 1; --i) {
    save[i] = cal(s[i], n);
    bit.modify(save[i], save[i]);
    vis[save[i]] = 1;
  }
  for (int i = 1; i <= n; ++i) cout << save[i] << " ";
  cout << endl;
  return 0;
}
inline ll read() {
  ll x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}

```