# Restore Permutation

An array of integers $p_{1},p_{2}, \ldots,p_{n}$ 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.

Input

The first line contains a single integer $n$ ($1 \le n \le 2 \cdot 10^{5}$) — the size of the permutation.

The second line contains $n$ integers $s_{1}, s_{2}, \ldots, s_{n}$ ($0 \le s_{i} \le \frac{n(n-1)}{2}$).

It is guaranteed that the array $s$ corresponds to a valid permutation of length $n$.

Output

Print $n$ integers $p_{1}, p_{2}, \ldots, p_{n}$ — 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 $s_5 = p_1 + p_2 + p_3 + p_4 = 10$.

Solution:

#include <bits/stdc++.h>

using namespace std;

template <typename T>
class fenwick {
public:
vector<T> fenw;
int n;

fenwick(int _n) : n(_n) {
fenw.resize(n);
}

void modify(int x, T v) {
while (x < n) {
fenw[x] += v;
x |= (x + 1);
}
}

T get(int x) {
T v{};
while (x >= 0) {
v += fenw[x];
x = (x & (x + 1)) - 1;
}
return v;
}
};

int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
vector<long long> s(n);
for (int i = 0; i < n; i++) {
cin >> s[i];
}
fenwick<long long> fenw(n);
for (int i = 0; i < n; i++) {
fenw.modify(i, i + 1);
}
vector<int> a(n);
for (int i = n - 1; i >= 0; i--) {
int low = 0, high = n - 1;
while (low < high) {
int mid = (low + high) >> 1;
if (fenw.get(mid) > s[i]) {
high = mid;
} else {
low = mid + 1;
}
}
a[i] = low + 1;
fenw.modify(low, -low - 1);
}
for (int i = 0; i < n; i++) {
if (i > 0) {
cout << " ";
}
cout << a[i];
}
cout << '\n';
return 0;
}