#include <bits/stdc++.h>
using namespace std;
int n;
long long IM = -999999999999;
typedef struct nde {
  int ll;
  int rl;
  long long val;
  int pos;
  long long prv;
  struct nde* l;
  struct nde* r;
  struct nde* pr;
} prt;
prt* getnew() {
  prt* p = new prt();
  p->prv = 0;
  return p;
}
prt* createTree(long long ar[], prt* pps[], int s, int e, prt* pr) {
  prt* rt = getnew();
  rt->ll = s;
  rt->rl = e;
  rt->pr = pr;
  rt->l = 0;
  rt->r = 0;
  if (s == e) {
    rt->pos = s;
    rt->val = ar[s];
    pps[s] = rt;
    return rt;
  } else {
    prt* t = createTree(ar, pps, s, (s + e) / 2, rt);
    prt* tt = createTree(ar, pps, 1 + ((s + e) / 2), e, rt);
    rt->l = t;
    rt->r = tt;
    if (t->val == tt->val) {
      rt->val = t->val;
      rt->pos = tt->pos;
    } else if (t->val < tt->val) {
      rt->val = t->val;
      rt->pos = t->pos;
    } else {
      rt->val = tt->val;
      rt->pos = tt->pos;
    }
  }
  return rt;
}
void uptd(int k, int rr, long long u, prt* p) {
  if (p->ll >= k && p->rl <= rr) {
    p->prv += u;
    return;
  }
  p->l->prv += p->prv;
  p->r->prv += p->prv;
  p->prv = 0;
  if (p->l->rl >= k && p->l->ll <= k) {
    uptd(k, rr, u, p->l);
  } else if (p->l->ll <= rr && p->l->rl >= rr) {
    uptd(k, rr, u, p->l);
  } else if (p->l->ll >= k && p->l->rl <= rr) {
    uptd(k, rr, u, p->l);
  } else {
  }
  if (p->r->rl >= k && p->r->ll <= k) {
    uptd(k, rr, u, p->r);
  } else if (p->r->ll <= rr && p->r->rl >= rr) {
    uptd(k, rr, u, p->r);
  } else if (p->r->ll >= k && p->r->rl <= rr) {
    uptd(k, rr, u, p->r);
  } else {
  }
  if ((p->l->val - p->l->prv) < (p->r->val - p->r->prv)) {
    p->val = p->l->val - p->l->prv;
    p->pos = p->l->pos;
  } else {
    p->val = p->r->val - p->r->prv;
    p->pos = p->r->pos;
  }
  return;
}
void inactive(int k, prt* root) { uptd(k, k, IM, root); }
int main() {
  int u;
  cin >> n;
  long long ar[n + 1];
  prt* pps[n + 1];
  int i = 1;
  while (i <= n) {
    cin >> ar[i];
    i++;
  }
  prt* root = createTree(ar, pps, 1, n, 0);
  i = 1;
  int t;
  while (i <= n) {
    t = root->pos;
    ar[t] = i;
    inactive(t, root);
    if (t < n) uptd(t + 1, n, i, root);
    i++;
  }
  i = 1;
  while (i <= n) {
    cout << ar[i] << " ";
    i++;
  }
  cout << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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;
int n;
long long IM = -999999999999;
typedef struct nde {
  int ll;
  int rl;
  long long val;
  int pos;
  long long prv;
  struct nde* l;
  struct nde* r;
  struct nde* pr;
} prt;
prt* getnew() {
  prt* p = new prt();
  p->prv = 0;
  return p;
}
prt* createTree(long long ar[], prt* pps[], int s, int e, prt* pr) {
  prt* rt = getnew();
  rt->ll = s;
  rt->rl = e;
  rt->pr = pr;
  rt->l = 0;
  rt->r = 0;
  if (s == e) {
    rt->pos = s;
    rt->val = ar[s];
    pps[s] = rt;
    return rt;
  } else {
    prt* t = createTree(ar, pps, s, (s + e) / 2, rt);
    prt* tt = createTree(ar, pps, 1 + ((s + e) / 2), e, rt);
    rt->l = t;
    rt->r = tt;
    if (t->val == tt->val) {
      rt->val = t->val;
      rt->pos = tt->pos;
    } else if (t->val < tt->val) {
      rt->val = t->val;
      rt->pos = t->pos;
    } else {
      rt->val = tt->val;
      rt->pos = tt->pos;
    }
  }
  return rt;
}
void uptd(int k, int rr, long long u, prt* p) {
  if (p->ll >= k && p->rl <= rr) {
    p->prv += u;
    return;
  }
  p->l->prv += p->prv;
  p->r->prv += p->prv;
  p->prv = 0;
  if (p->l->rl >= k && p->l->ll <= k) {
    uptd(k, rr, u, p->l);
  } else if (p->l->ll <= rr && p->l->rl >= rr) {
    uptd(k, rr, u, p->l);
  } else if (p->l->ll >= k && p->l->rl <= rr) {
    uptd(k, rr, u, p->l);
  } else {
  }
  if (p->r->rl >= k && p->r->ll <= k) {
    uptd(k, rr, u, p->r);
  } else if (p->r->ll <= rr && p->r->rl >= rr) {
    uptd(k, rr, u, p->r);
  } else if (p->r->ll >= k && p->r->rl <= rr) {
    uptd(k, rr, u, p->r);
  } else {
  }
  if ((p->l->val - p->l->prv) < (p->r->val - p->r->prv)) {
    p->val = p->l->val - p->l->prv;
    p->pos = p->l->pos;
  } else {
    p->val = p->r->val - p->r->prv;
    p->pos = p->r->pos;
  }
  return;
}
void inactive(int k, prt* root) { uptd(k, k, IM, root); }
int main() {
  int u;
  cin >> n;
  long long ar[n + 1];
  prt* pps[n + 1];
  int i = 1;
  while (i <= n) {
    cin >> ar[i];
    i++;
  }
  prt* root = createTree(ar, pps, 1, n, 0);
  i = 1;
  int t;
  while (i <= n) {
    t = root->pos;
    ar[t] = i;
    inactive(t, root);
    if (t < n) uptd(t + 1, n, i, root);
    i++;
  }
  i = 1;
  while (i <= n) {
    cout << ar[i] << " ";
    i++;
  }
  cout << endl;
  return 0;
}

```