#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
class {
 public:
  pair<long long, long long> T[200009 << 2];
  long long lz[200009 << 2];
  long long n, f[200009], ans[200009];
  void solve() {
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> f[i];
    build(1, 1, n);
    for (int i = 1; i <= n; ++i) {
      auto res = query(1, 1, n, 1, n);
      ans[res.second] = i;
      update(1, 1, n, res.second + 1, n, -i);
      update(1, 1, n, res.second, res.second, 1e18);
    }
    for (int i = 1; i <= n; ++i) cout << ans[i] << " ";
  }
  void build(int rt, int l, int r) {
    lz[rt] = 0;
    if (l == r) {
      T[rt] = {f[l], l};
      return;
    }
    int mid = l + r >> 1;
    build(rt << 1, l, mid);
    build(rt << 1 | 1, mid + 1, r);
    up(rt);
  }
  void up(int rt) {
    if (T[rt << 1].first == T[rt << 1 | 1].first) {
      T[rt] = T[rt << 1 | 1];
    } else {
      T[rt] = min(T[rt << 1], T[rt << 1 | 1]);
    }
  }
  void update(int rt, int l, int r, int ll, int rr, long long v) {
    if (ll > rr) return;
    if (ll <= l && rr >= r) {
      lz[rt] += v;
      T[rt].first += v;
      return;
    }
    int mid = l + r >> 1;
    down(rt);
    if (ll <= mid) update(rt << 1, l, mid, ll, rr, v);
    if (rr > mid) update(rt << 1 | 1, mid + 1, r, ll, rr, v);
    up(rt);
  }
  pair<long long, long long> query(int rt, int l, int r, int ll, int rr) {
    if (ll <= l && rr >= r) {
      return T[rt];
    }
    int mid = l + r >> 1;
    down(rt);
    pair<long long, long long> res = {1e18, 1e18};
    if (ll <= mid) {
      pair<long long, long long> ret = query(rt << 1, l, mid, ll, rr);
      if (ret.first == res.first)
        res.second = ret.second;
      else
        res = min(res, ret);
    }
    if (rr > mid) {
      pair<long long, long long> ret = query(rt << 1 | 1, mid + 1, r, ll, rr);
      if (ret.first == res.first)
        res.second = ret.second;
      else
        res = min(res, ret);
    }
    return res;
  }
  void down(int rt) {
    if (lz[rt]) {
      T[rt << 1].first += lz[rt];
      lz[rt << 1] += lz[rt];
      T[rt << 1 | 1].first += lz[rt];
      lz[rt << 1 | 1] += lz[rt];
      lz[rt] = 0;
    }
  }
} NSPACE;
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  NSPACE.solve();
}

### 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 MOD = 1e9 + 7;
class {
 public:
  pair<long long, long long> T[200009 << 2];
  long long lz[200009 << 2];
  long long n, f[200009], ans[200009];
  void solve() {
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> f[i];
    build(1, 1, n);
    for (int i = 1; i <= n; ++i) {
      auto res = query(1, 1, n, 1, n);
      ans[res.second] = i;
      update(1, 1, n, res.second + 1, n, -i);
      update(1, 1, n, res.second, res.second, 1e18);
    }
    for (int i = 1; i <= n; ++i) cout << ans[i] << " ";
  }
  void build(int rt, int l, int r) {
    lz[rt] = 0;
    if (l == r) {
      T[rt] = {f[l], l};
      return;
    }
    int mid = l + r >> 1;
    build(rt << 1, l, mid);
    build(rt << 1 | 1, mid + 1, r);
    up(rt);
  }
  void up(int rt) {
    if (T[rt << 1].first == T[rt << 1 | 1].first) {
      T[rt] = T[rt << 1 | 1];
    } else {
      T[rt] = min(T[rt << 1], T[rt << 1 | 1]);
    }
  }
  void update(int rt, int l, int r, int ll, int rr, long long v) {
    if (ll > rr) return;
    if (ll <= l && rr >= r) {
      lz[rt] += v;
      T[rt].first += v;
      return;
    }
    int mid = l + r >> 1;
    down(rt);
    if (ll <= mid) update(rt << 1, l, mid, ll, rr, v);
    if (rr > mid) update(rt << 1 | 1, mid + 1, r, ll, rr, v);
    up(rt);
  }
  pair<long long, long long> query(int rt, int l, int r, int ll, int rr) {
    if (ll <= l && rr >= r) {
      return T[rt];
    }
    int mid = l + r >> 1;
    down(rt);
    pair<long long, long long> res = {1e18, 1e18};
    if (ll <= mid) {
      pair<long long, long long> ret = query(rt << 1, l, mid, ll, rr);
      if (ret.first == res.first)
        res.second = ret.second;
      else
        res = min(res, ret);
    }
    if (rr > mid) {
      pair<long long, long long> ret = query(rt << 1 | 1, mid + 1, r, ll, rr);
      if (ret.first == res.first)
        res.second = ret.second;
      else
        res = min(res, ret);
    }
    return res;
  }
  void down(int rt) {
    if (lz[rt]) {
      T[rt << 1].first += lz[rt];
      lz[rt << 1] += lz[rt];
      T[rt << 1 | 1].first += lz[rt];
      lz[rt << 1 | 1] += lz[rt];
      lz[rt] = 0;
    }
  }
} NSPACE;
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  NSPACE.solve();
}

```