#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 2e5 + 10;
const long long INF = 0x3f3f3f3f;
const long long LLINF = 0x3f3f3f3f3f3f3f3f;
const long long MOD = 1e9 + 7;
long long segtree[MAXN << 2];
long long lazy[MAXN << 2];
long long arr[MAXN];
long long ans[MAXN];
void pushup(long long rt) {
  segtree[rt] = min(segtree[rt << 1], segtree[rt << 1 | 1]);
}
void pushdown(long long rt) {
  if (lazy[rt]) {
    segtree[rt << 1] = max(0ll, segtree[rt << 1] - lazy[rt]);
    segtree[rt << 1 | 1] = max(0ll, segtree[rt << 1 | 1] - lazy[rt]);
    lazy[rt << 1] += lazy[rt];
    lazy[rt << 1 | 1] += lazy[rt];
    lazy[rt] = 0;
  }
}
void build(long long l, long long r, long long rt) {
  lazy[rt] = 0;
  if (l == r) {
    segtree[rt] = arr[l];
    return;
  }
  long long m = l + r >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  pushup(rt);
}
long long query(long long l, long long r, long long rt) {
  if (l == r) {
    return l;
  }
  long long m = l + r >> 1;
  pushdown(rt);
  if (!segtree[rt << 1 | 1])
    return query(m + 1, r, rt << 1 | 1);
  else
    return query(l, m, rt << 1);
}
void update(long long l, long long r, long long L, long long R, long long C,
            long long rt) {
  if (L <= l && r <= R) {
    segtree[rt] = max(0ll, segtree[rt] - C);
    lazy[rt] += C;
    return;
  }
  long long m = l + r >> 1;
  pushdown(rt);
  if (L <= m) update(l, m, L, R, C, rt << 1);
  if (R > m) update(m + 1, r, L, R, C, rt << 1 | 1);
  pushup(rt);
}
void update(long long l, long long r, long long pos, long long rt) {
  if (l == r && l == pos) {
    segtree[rt] = LLINF;
    return;
  }
  long long m = l + r >> 1;
  pushdown(rt);
  if (pos <= m)
    update(l, m, pos, rt << 1);
  else
    update(m + 1, r, pos, rt << 1 | 1);
  pushup(rt);
}
signed main() {
  long long n;
  scanf("%lld", &n);
  for (long long i = 1; i <= n; ++i) {
    scanf("%lld", &arr[i]);
  }
  build(1, n, 1);
  long long now = 0;
  for (long long i = n; i > 0; --i) {
    long long index = query(1, n, 1);
    ans[index] = ++now;
    if (index + 1 <= n) update(1, n, index + 1, n, now, 1);
    update(1, n, index, 1);
  }
  for (long long i = 1; i <= n; ++i) printf("%lld ", ans[i]);
  return 0;
}

### Prompt

Please formulate 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;
const long long MAXN = 2e5 + 10;
const long long INF = 0x3f3f3f3f;
const long long LLINF = 0x3f3f3f3f3f3f3f3f;
const long long MOD = 1e9 + 7;
long long segtree[MAXN << 2];
long long lazy[MAXN << 2];
long long arr[MAXN];
long long ans[MAXN];
void pushup(long long rt) {
  segtree[rt] = min(segtree[rt << 1], segtree[rt << 1 | 1]);
}
void pushdown(long long rt) {
  if (lazy[rt]) {
    segtree[rt << 1] = max(0ll, segtree[rt << 1] - lazy[rt]);
    segtree[rt << 1 | 1] = max(0ll, segtree[rt << 1 | 1] - lazy[rt]);
    lazy[rt << 1] += lazy[rt];
    lazy[rt << 1 | 1] += lazy[rt];
    lazy[rt] = 0;
  }
}
void build(long long l, long long r, long long rt) {
  lazy[rt] = 0;
  if (l == r) {
    segtree[rt] = arr[l];
    return;
  }
  long long m = l + r >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  pushup(rt);
}
long long query(long long l, long long r, long long rt) {
  if (l == r) {
    return l;
  }
  long long m = l + r >> 1;
  pushdown(rt);
  if (!segtree[rt << 1 | 1])
    return query(m + 1, r, rt << 1 | 1);
  else
    return query(l, m, rt << 1);
}
void update(long long l, long long r, long long L, long long R, long long C,
            long long rt) {
  if (L <= l && r <= R) {
    segtree[rt] = max(0ll, segtree[rt] - C);
    lazy[rt] += C;
    return;
  }
  long long m = l + r >> 1;
  pushdown(rt);
  if (L <= m) update(l, m, L, R, C, rt << 1);
  if (R > m) update(m + 1, r, L, R, C, rt << 1 | 1);
  pushup(rt);
}
void update(long long l, long long r, long long pos, long long rt) {
  if (l == r && l == pos) {
    segtree[rt] = LLINF;
    return;
  }
  long long m = l + r >> 1;
  pushdown(rt);
  if (pos <= m)
    update(l, m, pos, rt << 1);
  else
    update(m + 1, r, pos, rt << 1 | 1);
  pushup(rt);
}
signed main() {
  long long n;
  scanf("%lld", &n);
  for (long long i = 1; i <= n; ++i) {
    scanf("%lld", &arr[i]);
  }
  build(1, n, 1);
  long long now = 0;
  for (long long i = n; i > 0; --i) {
    long long index = query(1, n, 1);
    ans[index] = ++now;
    if (index + 1 <= n) update(1, n, index + 1, n, now, 1);
    update(1, n, index, 1);
  }
  for (long long i = 1; i <= n; ++i) printf("%lld ", ans[i]);
  return 0;
}

```