#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll n;
ll a[200001];
ll sum[200001];
ll bit[200001];
ll p[200001];
template <typename T>
inline T read() {
  T x = 0;
  T multiplier = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') {
      multiplier = -1;
    }
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch & 15);
    ch = getchar();
  }
  return x * multiplier;
}
inline ll query(ll idx) {
  ll ans = 0;
  for (; idx; idx -= idx & -idx) {
    ans += bit[idx];
  }
  return ans;
}
inline void add(ll n, ll idx, ll val) {
  for (; idx <= n; idx += idx & -idx) {
    bit[idx] += val;
  }
}
inline ll get_idx(ll x) {
  ll l = 1, r = n, mid;
  ll idx;
  while (l <= r) {
    mid = (l + r + 1) >> 1;
    if (query(mid) <= x) {
      l = mid + 1;
      idx = mid;
    } else {
      r = mid - 1;
    }
  }
  return idx;
}
int main() {
  n = read<ll>();
  for (ll i = 1; i <= n; i++) {
    a[i] = read<ll>();
  }
  for (ll i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + i - 1;
  }
  for (ll i = 1; i <= n; i++) {
    bit[i] = sum[i] - sum[i - (i & -i)];
  }
  for (ll i = n; i >= 1; i--) {
    ll x = get_idx(a[i]);
    p[i] = x;
    add(n, x + 1, -x);
  }
  for (ll i = 1; i <= n; i++) {
    printf("%lld ", p[i]);
  }
  puts("");
  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;
using ll = long long;
ll n;
ll a[200001];
ll sum[200001];
ll bit[200001];
ll p[200001];
template <typename T>
inline T read() {
  T x = 0;
  T multiplier = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') {
      multiplier = -1;
    }
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch & 15);
    ch = getchar();
  }
  return x * multiplier;
}
inline ll query(ll idx) {
  ll ans = 0;
  for (; idx; idx -= idx & -idx) {
    ans += bit[idx];
  }
  return ans;
}
inline void add(ll n, ll idx, ll val) {
  for (; idx <= n; idx += idx & -idx) {
    bit[idx] += val;
  }
}
inline ll get_idx(ll x) {
  ll l = 1, r = n, mid;
  ll idx;
  while (l <= r) {
    mid = (l + r + 1) >> 1;
    if (query(mid) <= x) {
      l = mid + 1;
      idx = mid;
    } else {
      r = mid - 1;
    }
  }
  return idx;
}
int main() {
  n = read<ll>();
  for (ll i = 1; i <= n; i++) {
    a[i] = read<ll>();
  }
  for (ll i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + i - 1;
  }
  for (ll i = 1; i <= n; i++) {
    bit[i] = sum[i] - sum[i - (i & -i)];
  }
  for (ll i = n; i >= 1; i--) {
    ll x = get_idx(a[i]);
    p[i] = x;
    add(n, x + 1, -x);
  }
  for (ll i = 1; i <= n; i++) {
    printf("%lld ", p[i]);
  }
  puts("");
  return 0;
}

```