#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 10;
const long long INF = 0x3f3f3f3f3f3f3f3f;
long long n, tag[N << 2], ans[N], pos;
long long a[N], val[N << 2];
inline void pushup(long long o) { val[o] = min(val[o << 1], val[o << 1 | 1]); }
inline void apply(long long o, long long x) { val[o] += x, tag[o] += x; }
inline void pushdown(long long o) {
  if (tag[o]) {
    apply(o << 1, tag[o]);
    apply(o << 1 | 1, tag[o]);
    tag[o] = 0;
  }
}
inline void build(long long o, long long l, long long r) {
  if (l == r) {
    val[o] = a[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build(o << 1, l, mid), build(o << 1 | 1, mid + 1, r);
  pushup(o);
}
inline void change(long long o, long long l, long long r, long long x,
                   long long y, long long z) {
  if (x <= l && r <= y) {
    apply(o, z);
    return;
  }
  long long mid = (l + r) >> 1;
  pushdown(o);
  if (x <= mid) change(o << 1, l, mid, x, y, z);
  if (y > mid) change(o << 1 | 1, mid + 1, r, x, y, z);
  pushup(o);
}
inline long long last_zero(long long o, long long l, long long r) {
  if (l == r) return l;
  pushdown(o);
  long long mid = (l + r) >> 1;
  if (val[o << 1 | 1] == 0)
    return last_zero(o << 1 | 1, mid + 1, r);
  else
    return last_zero(o << 1, l, mid);
}
inline long long query_min(long long o, long long l, long long r, long long x,
                           long long y) {
  if (l == r) return val[o];
  pushdown(o);
  long long mid = (l + r) >> 1;
  long long res = INF;
  if (x <= mid)
    res = min(res, query_min(o << 1, l, mid, x, y));
  else if (y > mid)
    res = min(res, query_min(o << 1 | 1, mid + 1, r, x, y));
  return res;
}
signed main() {
  cin >> n;
  for (register long long i = (1); i <= (n); ++i) cin >> a[i];
  memset(val, INF, sizeof(val));
  build(1, 1, n);
  for (register long long i = (1); i <= (n); ++i) {
    pos = last_zero(1, 1, n);
    if (pos == 1) cerr << i << ' ' << pos << endl;
    ans[pos] = i;
    change(1, 1, n, pos, pos, INF);
    change(1, 1, n, pos, n, -1ll * i);
  }
  for (register 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;
const long long N = 2e5 + 10;
const long long INF = 0x3f3f3f3f3f3f3f3f;
long long n, tag[N << 2], ans[N], pos;
long long a[N], val[N << 2];
inline void pushup(long long o) { val[o] = min(val[o << 1], val[o << 1 | 1]); }
inline void apply(long long o, long long x) { val[o] += x, tag[o] += x; }
inline void pushdown(long long o) {
  if (tag[o]) {
    apply(o << 1, tag[o]);
    apply(o << 1 | 1, tag[o]);
    tag[o] = 0;
  }
}
inline void build(long long o, long long l, long long r) {
  if (l == r) {
    val[o] = a[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build(o << 1, l, mid), build(o << 1 | 1, mid + 1, r);
  pushup(o);
}
inline void change(long long o, long long l, long long r, long long x,
                   long long y, long long z) {
  if (x <= l && r <= y) {
    apply(o, z);
    return;
  }
  long long mid = (l + r) >> 1;
  pushdown(o);
  if (x <= mid) change(o << 1, l, mid, x, y, z);
  if (y > mid) change(o << 1 | 1, mid + 1, r, x, y, z);
  pushup(o);
}
inline long long last_zero(long long o, long long l, long long r) {
  if (l == r) return l;
  pushdown(o);
  long long mid = (l + r) >> 1;
  if (val[o << 1 | 1] == 0)
    return last_zero(o << 1 | 1, mid + 1, r);
  else
    return last_zero(o << 1, l, mid);
}
inline long long query_min(long long o, long long l, long long r, long long x,
                           long long y) {
  if (l == r) return val[o];
  pushdown(o);
  long long mid = (l + r) >> 1;
  long long res = INF;
  if (x <= mid)
    res = min(res, query_min(o << 1, l, mid, x, y));
  else if (y > mid)
    res = min(res, query_min(o << 1 | 1, mid + 1, r, x, y));
  return res;
}
signed main() {
  cin >> n;
  for (register long long i = (1); i <= (n); ++i) cin >> a[i];
  memset(val, INF, sizeof(val));
  build(1, 1, n);
  for (register long long i = (1); i <= (n); ++i) {
    pos = last_zero(1, 1, n);
    if (pos == 1) cerr << i << ' ' << pos << endl;
    ans[pos] = i;
    change(1, 1, n, pos, pos, INF);
    change(1, 1, n, pos, n, -1ll * i);
  }
  for (register long long i = (1); i <= (n); ++i) cout << ans[i] << ' ';
  return 0;
}

```