#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
const long long maxn = 2e5 + 10;
long long n, a[maxn], ans[maxn];
long long read() {
  long long 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;
}
void print(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
void write(long long x) {
  print(x);
  puts("");
}
struct Segment_Tree {
  struct Seg {
    long long mn, f;
  } tree[maxn << 2];
  void update(long long k) {
    tree[k].mn = min(tree[k << 1].mn, tree[k << 1 | 1].mn);
  }
  void down(long long k) {
    if (tree[k].f) {
      tree[k << 1].f += tree[k].f;
      tree[k << 1 | 1].f += tree[k].f;
      tree[k << 1].mn += tree[k].f;
      tree[k << 1 | 1].mn += tree[k].f;
      tree[k].f = 0;
    }
  }
  void build(long long k, long long l, long long r) {
    tree[k].f = 0;
    if (l == r) {
      tree[k].mn = a[l];
      return;
    }
    long long mid = (l + r) >> 1;
    build(k << 1, l, mid);
    build(k << 1 | 1, mid + 1, r);
    update(k);
  }
  void change(long long k, long long l, long long r, long long x, long long y,
              long long v) {
    if (x > y) return;
    if (x <= l && r <= y) {
      tree[k].mn += v;
      tree[k].f += v;
      return;
    }
    down(k);
    long long mid = (l + r) >> 1;
    if (x <= mid) change(k << 1, l, mid, x, y, v);
    if (mid < y) change(k << 1 | 1, mid + 1, r, x, y, v);
    update(k);
  }
  void erase(long long k, long long l, long long r, long long x) {
    if (l == r) {
      tree[k].mn = inf;
      tree[k].f = 0;
      return;
    }
    down(k);
    long long mid = (l + r) >> 1;
    if (x <= mid)
      erase(k << 1, l, mid, x);
    else
      erase(k << 1 | 1, mid + 1, r, x);
    update(k);
  }
  long long query(long long k, long long l, long long r) {
    if (l == r) return l;
    down(k);
    long long mid = (l + r) >> 1;
    if (!tree[k << 1 | 1].mn)
      return query(k << 1 | 1, mid + 1, r);
    else
      return query(k << 1, l, mid);
  }
} T;
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  T.build(1, 1, n);
  for (long long i = 1; i <= n; i++) {
    long long pos = T.query(1, 1, n);
    ans[pos] = i;
    T.change(1, 1, n, pos + 1, n, -i);
    T.erase(1, 1, n, pos);
  }
  for (long long i = 1; i <= n; i++) printf("%lld ", ans[i]);
  puts("");
  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 inf = 1e18;
const long long maxn = 2e5 + 10;
long long n, a[maxn], ans[maxn];
long long read() {
  long long 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;
}
void print(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
void write(long long x) {
  print(x);
  puts("");
}
struct Segment_Tree {
  struct Seg {
    long long mn, f;
  } tree[maxn << 2];
  void update(long long k) {
    tree[k].mn = min(tree[k << 1].mn, tree[k << 1 | 1].mn);
  }
  void down(long long k) {
    if (tree[k].f) {
      tree[k << 1].f += tree[k].f;
      tree[k << 1 | 1].f += tree[k].f;
      tree[k << 1].mn += tree[k].f;
      tree[k << 1 | 1].mn += tree[k].f;
      tree[k].f = 0;
    }
  }
  void build(long long k, long long l, long long r) {
    tree[k].f = 0;
    if (l == r) {
      tree[k].mn = a[l];
      return;
    }
    long long mid = (l + r) >> 1;
    build(k << 1, l, mid);
    build(k << 1 | 1, mid + 1, r);
    update(k);
  }
  void change(long long k, long long l, long long r, long long x, long long y,
              long long v) {
    if (x > y) return;
    if (x <= l && r <= y) {
      tree[k].mn += v;
      tree[k].f += v;
      return;
    }
    down(k);
    long long mid = (l + r) >> 1;
    if (x <= mid) change(k << 1, l, mid, x, y, v);
    if (mid < y) change(k << 1 | 1, mid + 1, r, x, y, v);
    update(k);
  }
  void erase(long long k, long long l, long long r, long long x) {
    if (l == r) {
      tree[k].mn = inf;
      tree[k].f = 0;
      return;
    }
    down(k);
    long long mid = (l + r) >> 1;
    if (x <= mid)
      erase(k << 1, l, mid, x);
    else
      erase(k << 1 | 1, mid + 1, r, x);
    update(k);
  }
  long long query(long long k, long long l, long long r) {
    if (l == r) return l;
    down(k);
    long long mid = (l + r) >> 1;
    if (!tree[k << 1 | 1].mn)
      return query(k << 1 | 1, mid + 1, r);
    else
      return query(k << 1, l, mid);
  }
} T;
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  T.build(1, 1, n);
  for (long long i = 1; i <= n; i++) {
    long long pos = T.query(1, 1, n);
    ans[pos] = i;
    T.change(1, 1, n, pos + 1, n, -i);
    T.erase(1, 1, n, pos);
  }
  for (long long i = 1; i <= n; i++) printf("%lld ", ans[i]);
  puts("");
  return 0;
}

```