#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 9;
long long s[maxn];
int p[maxn];
int vis[maxn];
int temp;
struct node {
  int l, r;
  long long minn, lazy;
} tree[maxn << 2];
void BT(int p, int l, int r) {
  tree[p].l = l, tree[p].r = r;
  if (l == r) {
    tree[p].minn = s[l];
    tree[p].lazy = 0;
    return;
  }
  int mid = (l + r) >> 1;
  BT(p << 1, l, mid);
  BT(p << 1 | 1, mid + 1, r);
  tree[p].minn = min(tree[p << 1].minn, tree[p << 1 | 1].minn);
  tree[p].lazy = 0;
}
void spread(int p) {
  tree[p << 1].lazy += tree[p].lazy;
  tree[p << 1].minn -= tree[p].lazy;
  tree[p << 1 | 1].lazy += tree[p].lazy;
  tree[p << 1 | 1].minn -= tree[p].lazy;
  tree[p].lazy = 0;
}
void Query(int p) {
  if (tree[p].l == tree[p].r) {
    tree[p].minn = 0x3f3f3f3f3f3f3f3f;
    temp = tree[p].l;
    return;
  }
  spread(p);
  if (tree[p << 1 | 1].minn == 0) {
    Query(p << 1 | 1);
  } else
    Query(p << 1);
  tree[p].minn = min(tree[p << 1].minn, tree[p << 1 | 1].minn);
}
void Update(int p, int l, int r, int v) {
  if (tree[p].l >= l && tree[p].r <= r) {
    tree[p].lazy += v;
    tree[p].minn -= v;
    return;
  }
  spread(p);
  int mid = (tree[p].l + tree[p].r) >> 1;
  if (l > mid) {
    Update(p << 1 | 1, l, r, v);
  } else if (r <= mid)
    Update(p << 1, l, r, v);
  else {
    Update(p << 1 | 1, l, r, v);
    Update(p << 1, l, r, v);
  }
  tree[p].minn = min(tree[p << 1].minn, tree[p << 1 | 1].minn);
}
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &s[i]);
    }
    BT(1, 1, n);
    for (int i = 1; i <= n; i++) {
      Query(1);
      p[temp] = i;
      if (temp < n) Update(1, temp + 1, n, i);
    }
    for (int i = 1; i <= n; i++) {
      if (i == 1)
        printf("%d", p[i]);
      else
        printf(" %d", p[i]);
    }
    printf("\n");
  }
}

### 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>
using namespace std;
const int maxn = 2e5 + 9;
long long s[maxn];
int p[maxn];
int vis[maxn];
int temp;
struct node {
  int l, r;
  long long minn, lazy;
} tree[maxn << 2];
void BT(int p, int l, int r) {
  tree[p].l = l, tree[p].r = r;
  if (l == r) {
    tree[p].minn = s[l];
    tree[p].lazy = 0;
    return;
  }
  int mid = (l + r) >> 1;
  BT(p << 1, l, mid);
  BT(p << 1 | 1, mid + 1, r);
  tree[p].minn = min(tree[p << 1].minn, tree[p << 1 | 1].minn);
  tree[p].lazy = 0;
}
void spread(int p) {
  tree[p << 1].lazy += tree[p].lazy;
  tree[p << 1].minn -= tree[p].lazy;
  tree[p << 1 | 1].lazy += tree[p].lazy;
  tree[p << 1 | 1].minn -= tree[p].lazy;
  tree[p].lazy = 0;
}
void Query(int p) {
  if (tree[p].l == tree[p].r) {
    tree[p].minn = 0x3f3f3f3f3f3f3f3f;
    temp = tree[p].l;
    return;
  }
  spread(p);
  if (tree[p << 1 | 1].minn == 0) {
    Query(p << 1 | 1);
  } else
    Query(p << 1);
  tree[p].minn = min(tree[p << 1].minn, tree[p << 1 | 1].minn);
}
void Update(int p, int l, int r, int v) {
  if (tree[p].l >= l && tree[p].r <= r) {
    tree[p].lazy += v;
    tree[p].minn -= v;
    return;
  }
  spread(p);
  int mid = (tree[p].l + tree[p].r) >> 1;
  if (l > mid) {
    Update(p << 1 | 1, l, r, v);
  } else if (r <= mid)
    Update(p << 1, l, r, v);
  else {
    Update(p << 1 | 1, l, r, v);
    Update(p << 1, l, r, v);
  }
  tree[p].minn = min(tree[p << 1].minn, tree[p << 1 | 1].minn);
}
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &s[i]);
    }
    BT(1, 1, n);
    for (int i = 1; i <= n; i++) {
      Query(1);
      p[temp] = i;
      if (temp < n) Update(1, temp + 1, n, i);
    }
    for (int i = 1; i <= n; i++) {
      if (i == 1)
        printf("%d", p[i]);
      else
        printf(" %d", p[i]);
    }
    printf("\n");
  }
}

```