#include <bits/stdc++.h>
using namespace std;
struct Node {
  long long l, r, rt, lz, minn, maxx;
} node[1008611];
long long a[200861], sum[400861];
long long ans[200861];
void build(long long l, long long r, long long rt) {
  long long mid = (l + r) / 2;
  node[rt].l = l, node[rt].r = r;
  node[rt].lz = 0;
  if (l == r) {
    node[rt].minn = sum[l];
    node[rt].maxx = sum[l];
    return;
  }
  build(l, mid, rt * 2);
  build(mid + 1, r, rt * 2 + 1);
  node[rt].minn = min(node[rt * 2].minn, node[rt * 2 + 1].minn);
  node[rt].maxx = max(node[rt * 2].maxx, node[rt * 2 + 1].maxx);
}
void push_down(long long rt) {
  if (node[rt].lz) {
    node[rt * 2].minn -= node[rt].lz;
    node[rt * 2 + 1].minn -= node[rt].lz;
    node[rt * 2].maxx -= node[rt].lz;
    node[rt * 2 + 1].maxx -= node[rt].lz;
    node[rt * 2].lz += node[rt].lz;
    node[rt * 2 + 1].lz += node[rt].lz;
    node[rt].lz = 0;
  }
}
long long query(long long v, long long rt) {
  if (node[rt].l == node[rt].r) {
    return node[rt].r + 1;
  }
  push_down(rt);
  if (node[rt * 2 + 1].minn <= v) {
    return query(v, rt * 2 + 1);
  } else if (node[rt * 2].maxx >= v) {
    return query(v, rt * 2);
  }
  node[rt].minn = min(node[rt * 2].minn, node[rt * 2 + 1].minn);
  node[rt].maxx = max(node[rt * 2].maxx, node[rt * 2 + 1].maxx);
}
void update(long long l, long long r, long long v, long long rt) {
  long long mid = (node[rt].l + node[rt].r) / 2;
  if (node[rt].l == l && node[rt].r == r) {
    node[rt].lz += v;
    node[rt].minn -= v;
    node[rt].maxx -= v;
    return;
  }
  push_down(rt);
  if (r <= mid) {
    update(l, r, v, rt * 2);
  } else if (l > mid) {
    update(l, r, v, rt * 2 + 1);
  } else {
    update(l, mid, v, rt * 2);
    update(mid + 1, r, v, rt * 2 + 1);
  }
  node[rt].minn = min(node[rt * 2].minn, node[rt * 2 + 1].minn);
  node[rt].maxx = max(node[rt * 2].maxx, node[rt * 2 + 1].maxx);
}
int main() {
  long long n, i, j;
  scanf("%lld", &n);
  sum[0] = 0;
  for (i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    sum[i] = sum[i - 1] + i;
  }
  build(0, n, 1);
  for (i = n; i >= 1; i--) {
    ans[i] = query(a[i], 1);
    update(ans[i], n, ans[i], 1);
  }
  for (i = 1; i <= n; i++) {
    if (i == n)
      printf("%lld\n", ans[i]);
    else if (i < n) {
      printf("%lld ", ans[i]);
    }
  }
  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;
struct Node {
  long long l, r, rt, lz, minn, maxx;
} node[1008611];
long long a[200861], sum[400861];
long long ans[200861];
void build(long long l, long long r, long long rt) {
  long long mid = (l + r) / 2;
  node[rt].l = l, node[rt].r = r;
  node[rt].lz = 0;
  if (l == r) {
    node[rt].minn = sum[l];
    node[rt].maxx = sum[l];
    return;
  }
  build(l, mid, rt * 2);
  build(mid + 1, r, rt * 2 + 1);
  node[rt].minn = min(node[rt * 2].minn, node[rt * 2 + 1].minn);
  node[rt].maxx = max(node[rt * 2].maxx, node[rt * 2 + 1].maxx);
}
void push_down(long long rt) {
  if (node[rt].lz) {
    node[rt * 2].minn -= node[rt].lz;
    node[rt * 2 + 1].minn -= node[rt].lz;
    node[rt * 2].maxx -= node[rt].lz;
    node[rt * 2 + 1].maxx -= node[rt].lz;
    node[rt * 2].lz += node[rt].lz;
    node[rt * 2 + 1].lz += node[rt].lz;
    node[rt].lz = 0;
  }
}
long long query(long long v, long long rt) {
  if (node[rt].l == node[rt].r) {
    return node[rt].r + 1;
  }
  push_down(rt);
  if (node[rt * 2 + 1].minn <= v) {
    return query(v, rt * 2 + 1);
  } else if (node[rt * 2].maxx >= v) {
    return query(v, rt * 2);
  }
  node[rt].minn = min(node[rt * 2].minn, node[rt * 2 + 1].minn);
  node[rt].maxx = max(node[rt * 2].maxx, node[rt * 2 + 1].maxx);
}
void update(long long l, long long r, long long v, long long rt) {
  long long mid = (node[rt].l + node[rt].r) / 2;
  if (node[rt].l == l && node[rt].r == r) {
    node[rt].lz += v;
    node[rt].minn -= v;
    node[rt].maxx -= v;
    return;
  }
  push_down(rt);
  if (r <= mid) {
    update(l, r, v, rt * 2);
  } else if (l > mid) {
    update(l, r, v, rt * 2 + 1);
  } else {
    update(l, mid, v, rt * 2);
    update(mid + 1, r, v, rt * 2 + 1);
  }
  node[rt].minn = min(node[rt * 2].minn, node[rt * 2 + 1].minn);
  node[rt].maxx = max(node[rt * 2].maxx, node[rt * 2 + 1].maxx);
}
int main() {
  long long n, i, j;
  scanf("%lld", &n);
  sum[0] = 0;
  for (i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    sum[i] = sum[i - 1] + i;
  }
  build(0, n, 1);
  for (i = n; i >= 1; i--) {
    ans[i] = query(a[i], 1);
    update(ans[i], n, ans[i], 1);
  }
  for (i = 1; i <= n; i++) {
    if (i == n)
      printf("%lld\n", ans[i]);
    else if (i < n) {
      printf("%lld ", ans[i]);
    }
  }
  return 0;
}

```