#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const long long inf = 1000000000000;
const int N = 200005;
const long long LLMAX = 2e18;
const long long MOD = 1000000009ll;
const double eps = 1e-8;
const int MAXN = 1e6 + 10;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int read() {
  int 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;
}
int n, ans[N];
long long a[N];
struct SegmentTree {
  int l, r, id;
  long long v, add;
} tree[N * 4];
void push_up(int i) {
  if (tree[i << 1 | 1].v == tree[i << 1].v) {
    if (tree[i << 1 | 1].id > tree[i << 1].id) {
      tree[i].id = tree[i << 1 | 1].id;
      tree[i].v = tree[i << 1 | 1].v;
    } else {
      tree[i].id = tree[i << 1].id;
      tree[i].v = tree[i << 1].v;
    }
  } else {
    if (tree[i << 1].v < tree[i << 1 | 1].v) {
      tree[i].id = tree[i << 1].id;
      tree[i].v = tree[i << 1].v;
    } else {
      tree[i].id = tree[i << 1 | 1].id;
      tree[i].v = tree[i << 1 | 1].v;
    }
  }
  return;
}
void build(int p, int l, int r) {
  tree[p].l = l, tree[p].r = r;
  if (l == r) {
    tree[p].v = a[l];
    tree[p].id = r;
    return;
  }
  int mid = (l + r) >> 1;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
  push_up(p);
}
void spread(int p) {
  if (tree[p].add) {
    tree[p << 1].add += tree[p].add;
    tree[p << 1 | 1].add += tree[p].add;
    tree[p << 1].v += tree[p].add;
    tree[p << 1 | 1].v += tree[p].add;
    tree[p].add = 0;
  }
}
void change(int p, int l, int r, long long d) {
  if (l <= tree[p].l && r >= tree[p].r) {
    tree[p].v += d;
    tree[p].add += d;
    return;
  }
  spread(p);
  int mid = (tree[p].l + tree[p].r) >> 1;
  if (l <= mid) change(p << 1, l, r, d);
  if (r > mid) change(p << 1 | 1, l, r, d);
  push_up(p);
}
int main(int argc, char const *argv[]) {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]);
  build(1, 1, n);
  for (int i = 1; i <= n; ++i) {
    ans[tree[1].id] = i;
    int d = tree[1].id;
    change(1, d, d, 10000000000000);
    change(1, d + 1, n, -i);
  }
  for (int i = 1; i <= n; ++i) {
    printf("%d%c", ans[i], " \n"[i == n]);
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const long long inf = 1000000000000;
const int N = 200005;
const long long LLMAX = 2e18;
const long long MOD = 1000000009ll;
const double eps = 1e-8;
const int MAXN = 1e6 + 10;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int read() {
  int 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;
}
int n, ans[N];
long long a[N];
struct SegmentTree {
  int l, r, id;
  long long v, add;
} tree[N * 4];
void push_up(int i) {
  if (tree[i << 1 | 1].v == tree[i << 1].v) {
    if (tree[i << 1 | 1].id > tree[i << 1].id) {
      tree[i].id = tree[i << 1 | 1].id;
      tree[i].v = tree[i << 1 | 1].v;
    } else {
      tree[i].id = tree[i << 1].id;
      tree[i].v = tree[i << 1].v;
    }
  } else {
    if (tree[i << 1].v < tree[i << 1 | 1].v) {
      tree[i].id = tree[i << 1].id;
      tree[i].v = tree[i << 1].v;
    } else {
      tree[i].id = tree[i << 1 | 1].id;
      tree[i].v = tree[i << 1 | 1].v;
    }
  }
  return;
}
void build(int p, int l, int r) {
  tree[p].l = l, tree[p].r = r;
  if (l == r) {
    tree[p].v = a[l];
    tree[p].id = r;
    return;
  }
  int mid = (l + r) >> 1;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
  push_up(p);
}
void spread(int p) {
  if (tree[p].add) {
    tree[p << 1].add += tree[p].add;
    tree[p << 1 | 1].add += tree[p].add;
    tree[p << 1].v += tree[p].add;
    tree[p << 1 | 1].v += tree[p].add;
    tree[p].add = 0;
  }
}
void change(int p, int l, int r, long long d) {
  if (l <= tree[p].l && r >= tree[p].r) {
    tree[p].v += d;
    tree[p].add += d;
    return;
  }
  spread(p);
  int mid = (tree[p].l + tree[p].r) >> 1;
  if (l <= mid) change(p << 1, l, r, d);
  if (r > mid) change(p << 1 | 1, l, r, d);
  push_up(p);
}
int main(int argc, char const *argv[]) {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]);
  build(1, 1, n);
  for (int i = 1; i <= n; ++i) {
    ans[tree[1].id] = i;
    int d = tree[1].id;
    change(1, d, d, 10000000000000);
    change(1, d + 1, n, -i);
  }
  for (int i = 1; i <= n; ++i) {
    printf("%d%c", ans[i], " \n"[i == n]);
  }
  return 0;
}

```