#include <bits/stdc++.h>
using namespace std;
long long read() {
  char ch = getchar();
  long long x = 0, ff = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') ff = -ff;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * ff;
}
void write(long long aa) {
  if (aa < 0) putchar('-'), aa = -aa;
  if (aa > 9) write(aa / 10);
  putchar(aa % 10 + '0');
  return;
}
long long n;
long long a[200005];
long long tt[800005], tts[800005], lz[800005];
void up(long long rt) {
  tt[rt] = min(tt[rt << 1], tt[rt << 1 | 1]);
  if (tt[rt << 1] >= tt[rt << 1 | 1])
    tts[rt] = tts[rt << 1 | 1];
  else
    tts[rt] = tts[rt << 1];
  return;
}
void bt(long long rt, long long ll, long long rr) {
  if (ll == rr) {
    tt[rt] = read(), tts[rt] = ll;
    return;
  }
  long long mid = (ll + rr) >> 1;
  bt(rt << 1, ll, mid);
  bt(rt << 1 | 1, mid + 1, rr);
  up(rt);
  return;
}
void push(long long rt) {
  if (!lz[rt]) return;
  tt[rt << 1] += lz[rt];
  tt[rt << 1 | 1] += lz[rt];
  lz[rt << 1] += lz[rt];
  lz[rt << 1 | 1] += lz[rt];
  lz[rt] = 0;
  return;
}
void update(long long rt, long long ll, long long rr, long long L, long long R,
            long long kk) {
  if (ll == L && rr == R) {
    tt[rt] += kk;
    lz[rt] += kk;
    return;
  }
  long long mid = (ll + rr) >> 1;
  push(rt);
  if (R <= mid)
    update(rt << 1, ll, mid, L, R, kk);
  else if (L > mid)
    update(rt << 1 | 1, mid + 1, rr, L, R, kk);
  else
    update(rt << 1, ll, mid, L, mid, kk),
        update(rt << 1 | 1, mid + 1, rr, mid + 1, R, kk);
  up(rt);
  return;
}
int main() {
  n = read();
  bt(1, 1, n);
  for (long long i = 1; i <= n; ++i) {
    long long now = tts[1];
    a[now] = i;
    update(1, 1, n, now, now, n * n);
    if (now < n) update(1, 1, n, now + 1, n, -i);
  }
  for (long long i = 1; i <= n; ++i, putchar(' ')) write(a[i]);
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
long long read() {
  char ch = getchar();
  long long x = 0, ff = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') ff = -ff;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * ff;
}
void write(long long aa) {
  if (aa < 0) putchar('-'), aa = -aa;
  if (aa > 9) write(aa / 10);
  putchar(aa % 10 + '0');
  return;
}
long long n;
long long a[200005];
long long tt[800005], tts[800005], lz[800005];
void up(long long rt) {
  tt[rt] = min(tt[rt << 1], tt[rt << 1 | 1]);
  if (tt[rt << 1] >= tt[rt << 1 | 1])
    tts[rt] = tts[rt << 1 | 1];
  else
    tts[rt] = tts[rt << 1];
  return;
}
void bt(long long rt, long long ll, long long rr) {
  if (ll == rr) {
    tt[rt] = read(), tts[rt] = ll;
    return;
  }
  long long mid = (ll + rr) >> 1;
  bt(rt << 1, ll, mid);
  bt(rt << 1 | 1, mid + 1, rr);
  up(rt);
  return;
}
void push(long long rt) {
  if (!lz[rt]) return;
  tt[rt << 1] += lz[rt];
  tt[rt << 1 | 1] += lz[rt];
  lz[rt << 1] += lz[rt];
  lz[rt << 1 | 1] += lz[rt];
  lz[rt] = 0;
  return;
}
void update(long long rt, long long ll, long long rr, long long L, long long R,
            long long kk) {
  if (ll == L && rr == R) {
    tt[rt] += kk;
    lz[rt] += kk;
    return;
  }
  long long mid = (ll + rr) >> 1;
  push(rt);
  if (R <= mid)
    update(rt << 1, ll, mid, L, R, kk);
  else if (L > mid)
    update(rt << 1 | 1, mid + 1, rr, L, R, kk);
  else
    update(rt << 1, ll, mid, L, mid, kk),
        update(rt << 1 | 1, mid + 1, rr, mid + 1, R, kk);
  up(rt);
  return;
}
int main() {
  n = read();
  bt(1, 1, n);
  for (long long i = 1; i <= n; ++i) {
    long long now = tts[1];
    a[now] = i;
    update(1, 1, n, now, now, n * n);
    if (now < n) update(1, 1, n, now + 1, n, -i);
  }
  for (long long i = 1; i <= n; ++i, putchar(' ')) write(a[i]);
  return 0;
}

```