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.
Your task is to restore the permutation.
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; }