Leha is a usual student at 'The Usual University for Usual Students'. Sometimes he studies hard; at other times he plays truant and gets busy with other things besides academics. He has already studied at the university for N months. For the ith month (1 ≤ i ≤ N), he has received some non-negative integer grade A[i].
Now he wants to analyse his progress for some periods of his university education. An arbitrary period, defined by two positive integers L and R, begins at Leha's Lth month at the university and ends at the Rth. The analysis is performed via the following steps.

1. Write down all the grades for each month from L to R and sort them. Let's call the sorted list S.

2. Calculate the sum of squared differences of consecutive elements in S, that is, (S[2] - S[1])2 + (S[3] - S[2])2 + ... + (S[R-L+1] - S[R-L])2.

-----Input-----
The first line contains one integer N — the number of months Leha has already studied at the university.
The second line contains N integers — list A of Leha's grades.
The third line contains one integer M — the number of periods Leha is interested in analyzing.
Each of the following M lines contain two integers L and R describing each period.

-----Output-----
For each query, output one integer — the result of the progress analysis for the corresponding period.

-----Constraints-----
- 1 ≤ N, M ≤ 5*104
- 0 ≤ A[i] ≤ 106

-----Subtasks-----
- Subtask 1 (19 points) 1 ≤ N, M ≤ 200, time limit = 2 sec
- Subtask 2 (31 points) 1 ≤ N, M ≤ 10 000, time limit = 2 sec
- Subtask 3 (26 points) 0 ≤ A[i] ≤ 100, time limit = 5 sec
- Subtask 4 (24 points) no additional constraints, , time limit = 5 sec

-----Example-----
Input:5
1 3 2 4 5
5
1 5
1 4
2 4
3 3
3 5

Output:4
3
2
0
5

Explanation

The first query: sorted array looks like (1, 2, 3, 4, 5) and the answer is calculated as (2-1)2 + (3-2)2 + (4-3)2 + (5-4)2 = 4

The second query: sorted array looks like (1, 2, 3, 4) and the answer is calculated as (2-1)2 + (3-2)2 + (4-3)2 = 3

The last query: sorted array looks like (2, 4, 5) and the answer is calculated as (4-2)2 + (5-4)2 = 5

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'n,a,m=eval(input()),list(map(int,input().split())),eval(input())\nfor i__i in range(m):\n l,r= list(map(int,input().split()))\n l-=1\n Rev,_=a[l:r],0\n Rev.sort()\n d=len(Rev)\n for i_i in range(0,d-1):\n  _+=(Rev[i_i+1]-Rev[i_i])*(Rev[i_i+1]-Rev[i_i])\n print(_)\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 17735