#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e17;
struct node {
  long long ans, sub;
  node() : ans(0), sub(0){};
};
struct SegmentTree {
  long long N;
  vector<node> st;
  vector<long long> a;
  long long left(long long p) { return p << 1; }
  long long right(long long p) { return (p << 1) + 1; }
  void push(long long p) {
    st[left(p)].sub += st[p].sub;
    st[right(p)].sub += st[p].sub;
    st[left(p)].ans -= st[p].sub;
    st[right(p)].ans -= st[p].sub;
    st[p].sub = 0;
  }
  void merge(long long p) {
    st[p].ans = min(st[left(p)].ans, st[right(p)].ans);
  }
  void build(long long p, long long l, long long r) {
    if (l == r) {
      st[p].ans = a[l];
    } else {
      long long mid = l + r >> 1;
      build(left(p), l, mid);
      build(right(p), mid + 1, r);
      merge(p);
    }
  }
  SegmentTree(long long N, vector<long long>& a) : N(N), st(4 * N), a(a) {
    build(1, 0, N - 1);
  }
  void setElem(long long p, long long l, long long r, long long i) {
    if (l == r) {
      st[p].ans = inf;
    } else {
      long long mid = l + r >> 1;
      push(p);
      if (i > mid)
        setElem(right(p), mid + 1, r, i);
      else
        setElem(left(p), l, mid, i);
      merge(p);
    }
  }
  void update(long long p, long long l, long long r, long long i, long long j,
              long long val) {
    if (l == i && r == j) {
      st[p].sub += val;
      st[p].ans -= val;
    } else {
      long long mid = l + r >> 1;
      push(p);
      if (i > mid)
        update(right(p), mid + 1, r, i, j, val);
      else if (j <= mid)
        update(left(p), l, mid, i, j, val);
      else {
        update(left(p), l, mid, i, mid, val);
        update(right(p), mid + 1, r, mid + 1, j, val);
      }
      merge(p);
    }
  }
  void update(long long i, long long val) {
    if (i + 1 < N) update(1, 0, N - 1, i + 1, N - 1, val);
    setElem(1, 0, N - 1, i);
  }
  long long query(long long p, long long l, long long r) {
    if (l == r) {
      return l;
    } else {
      long long mid = l + r >> 1;
      push(p);
      if (st[right(p)].ans == 0)
        return query(right(p), mid + 1, r);
      else
        return query(left(p), l, mid);
    }
  }
  long long query() { return query(1, 0, N - 1); }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  vector<long long> s(n);
  for (long long i = 0; i < n; i++) cin >> s[i];
  vector<int> ans(n);
  SegmentTree st(n, s);
  for (long long i = 0; i < n; i++) {
    long long idx = st.query();
    ans[idx] = i + 1;
    st.update(idx, i + 1);
  }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}

### 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;
const long long inf = 1e17;
struct node {
  long long ans, sub;
  node() : ans(0), sub(0){};
};
struct SegmentTree {
  long long N;
  vector<node> st;
  vector<long long> a;
  long long left(long long p) { return p << 1; }
  long long right(long long p) { return (p << 1) + 1; }
  void push(long long p) {
    st[left(p)].sub += st[p].sub;
    st[right(p)].sub += st[p].sub;
    st[left(p)].ans -= st[p].sub;
    st[right(p)].ans -= st[p].sub;
    st[p].sub = 0;
  }
  void merge(long long p) {
    st[p].ans = min(st[left(p)].ans, st[right(p)].ans);
  }
  void build(long long p, long long l, long long r) {
    if (l == r) {
      st[p].ans = a[l];
    } else {
      long long mid = l + r >> 1;
      build(left(p), l, mid);
      build(right(p), mid + 1, r);
      merge(p);
    }
  }
  SegmentTree(long long N, vector<long long>& a) : N(N), st(4 * N), a(a) {
    build(1, 0, N - 1);
  }
  void setElem(long long p, long long l, long long r, long long i) {
    if (l == r) {
      st[p].ans = inf;
    } else {
      long long mid = l + r >> 1;
      push(p);
      if (i > mid)
        setElem(right(p), mid + 1, r, i);
      else
        setElem(left(p), l, mid, i);
      merge(p);
    }
  }
  void update(long long p, long long l, long long r, long long i, long long j,
              long long val) {
    if (l == i && r == j) {
      st[p].sub += val;
      st[p].ans -= val;
    } else {
      long long mid = l + r >> 1;
      push(p);
      if (i > mid)
        update(right(p), mid + 1, r, i, j, val);
      else if (j <= mid)
        update(left(p), l, mid, i, j, val);
      else {
        update(left(p), l, mid, i, mid, val);
        update(right(p), mid + 1, r, mid + 1, j, val);
      }
      merge(p);
    }
  }
  void update(long long i, long long val) {
    if (i + 1 < N) update(1, 0, N - 1, i + 1, N - 1, val);
    setElem(1, 0, N - 1, i);
  }
  long long query(long long p, long long l, long long r) {
    if (l == r) {
      return l;
    } else {
      long long mid = l + r >> 1;
      push(p);
      if (st[right(p)].ans == 0)
        return query(right(p), mid + 1, r);
      else
        return query(left(p), l, mid);
    }
  }
  long long query() { return query(1, 0, N - 1); }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  vector<long long> s(n);
  for (long long i = 0; i < n; i++) cin >> s[i];
  vector<int> ans(n);
  SegmentTree st(n, s);
  for (long long i = 0; i < n; i++) {
    long long idx = st.query();
    ans[idx] = i + 1;
    st.update(idx, i + 1);
  }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}

```