#include <bits/stdc++.h>
inline long long minn(long long a, long long b) {
  if (a < b) return a;
  return b;
}
const int MAXN = 2e5 + 5;
const long long INF = 1ll << 60;
int n;
long long mi[MAXN << 2], tag[MAXN << 2];
inline void pushdwn(int k) {
  mi[k << 1] -= tag[k];
  tag[k << 1] += tag[k];
  mi[k << 1 | 1] -= tag[k];
  tag[k << 1 | 1] += tag[k];
  tag[k] = 0;
}
inline void pushup(int k) { mi[k] = minn(mi[k << 1], mi[k << 1 | 1]); }
void bld(int k, int l, int r) {
  if (l == r) {
    scanf("%I64d", mi + k);
    return;
  }
  int mid = l + r >> 1;
  bld(k << 1, l, mid);
  bld(k << 1 | 1, mid + 1, r);
  pushup(k);
  return;
}
int qx, qy, qv;
int Query(int k, int l, int r) {
  if (l == r) return l;
  pushdwn(k);
  int mid = l + r >> 1;
  if (!mi[k << 1 | 1]) return Query(k << 1 | 1, mid + 1, r);
  return Query(k << 1, l, mid);
}
void Modify2(int k, int l, int r) {
  if (l == r) {
    if (mi[k] < 0) mi[k] = INF;
    return;
  }
  pushdwn(k);
  if (mi[k] < 0) {
    int mid = l + r >> 1;
    Modify2(k << 1, l, mid);
    Modify2(k << 1 | 1, mid + 1, r);
    pushup(k);
  }
  return;
}
void Modify(int k, int l, int r) {
  if (qx <= l && r <= qy) {
    mi[k] -= qv;
    tag[k] += qv;
    if (mi[k] < 0) {
      if (l == r) {
        mi[k] = INF;
        return;
      }
      Modify2(k, l, r);
    }
    return;
  }
  pushdwn(k);
  int mid = l + r >> 1;
  if (qx <= mid) Modify(k << 1, l, mid);
  if (mid < qy) Modify(k << 1 | 1, mid + 1, r);
  pushup(k);
  return;
}
int ans[MAXN];
int main() {
  scanf("%d", &n);
  bld(1, 1, n);
  for (int i = 1; i <= n; i++) {
    int t = Query(1, 1, n);
    ans[t] = i;
    qx = t, qy = n, qv = i;
    Modify(1, 1, n);
  }
  for (int i = 1; i <= n; i++) printf("%d%c", ans[i], i == n ? '\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>
inline long long minn(long long a, long long b) {
  if (a < b) return a;
  return b;
}
const int MAXN = 2e5 + 5;
const long long INF = 1ll << 60;
int n;
long long mi[MAXN << 2], tag[MAXN << 2];
inline void pushdwn(int k) {
  mi[k << 1] -= tag[k];
  tag[k << 1] += tag[k];
  mi[k << 1 | 1] -= tag[k];
  tag[k << 1 | 1] += tag[k];
  tag[k] = 0;
}
inline void pushup(int k) { mi[k] = minn(mi[k << 1], mi[k << 1 | 1]); }
void bld(int k, int l, int r) {
  if (l == r) {
    scanf("%I64d", mi + k);
    return;
  }
  int mid = l + r >> 1;
  bld(k << 1, l, mid);
  bld(k << 1 | 1, mid + 1, r);
  pushup(k);
  return;
}
int qx, qy, qv;
int Query(int k, int l, int r) {
  if (l == r) return l;
  pushdwn(k);
  int mid = l + r >> 1;
  if (!mi[k << 1 | 1]) return Query(k << 1 | 1, mid + 1, r);
  return Query(k << 1, l, mid);
}
void Modify2(int k, int l, int r) {
  if (l == r) {
    if (mi[k] < 0) mi[k] = INF;
    return;
  }
  pushdwn(k);
  if (mi[k] < 0) {
    int mid = l + r >> 1;
    Modify2(k << 1, l, mid);
    Modify2(k << 1 | 1, mid + 1, r);
    pushup(k);
  }
  return;
}
void Modify(int k, int l, int r) {
  if (qx <= l && r <= qy) {
    mi[k] -= qv;
    tag[k] += qv;
    if (mi[k] < 0) {
      if (l == r) {
        mi[k] = INF;
        return;
      }
      Modify2(k, l, r);
    }
    return;
  }
  pushdwn(k);
  int mid = l + r >> 1;
  if (qx <= mid) Modify(k << 1, l, mid);
  if (mid < qy) Modify(k << 1 | 1, mid + 1, r);
  pushup(k);
  return;
}
int ans[MAXN];
int main() {
  scanf("%d", &n);
  bld(1, 1, n);
  for (int i = 1; i <= n; i++) {
    int t = Query(1, 1, n);
    ans[t] = i;
    qx = t, qy = n, qv = i;
    Modify(1, 1, n);
  }
  for (int i = 1; i <= n; i++) printf("%d%c", ans[i], i == n ? '\n' : ' ');
  return 0;
}

```