Firdavs is living on planet F. There are $N$ cities (numbered $1$ through $N$) on this planet; let's denote the value of city $i$ by $v_i$. Firdavs can travel directly from each city to any other city. When he travels directly from city $x$ to city $y$, he needs to pay $f(x, y) = |v_y-v_x|+y-x$ coins (this number can be negative, in which case he receives $-f(x, y)$ coins).
Let's define a simple path from city $x$ to city $y$ with length $k \ge 1$ as a sequence of cities $a_1, a_2, \ldots, a_k$ such that all cities in this sequence are different, $a_1 = x$ and $a_k = y$. The cost of such a path is $\sum_{i=1}^{k-1} f(a_i, a_{i+1})$.
You need to answer some queries for Firdavs. In each query, you are given two cities $x$ and $y$, and you need to find the minimum cost of a simple path from city $x$ to city $y$. Then, you need to find the length of the longest simple path from $x$ to $y$ with this cost.

-----Input-----
- The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
- The first line of each test case contains two space-separated integers $N$ and $Q$.
- The second line contains $N$ space-separated integers $v_1, v_2, \ldots, v_N$.
- The following $Q$ lines describe queries. Each of these lines contains two space-separated integers $x$ and $y$.

-----Output-----
For each query, print a single line containing two space-separated integers ― the minimum cost and the maximum length.

-----Constraints-----
- $1 \le T \le 100$
- $1 \le N, Q \le 2 \cdot 10^5$
- $0 \le v_i \le 10^9$ for each valid $i$
- $1 \le x, y \le N$
- the sum of $N$ in all test cases does not exceed $5 \cdot 10^5$
- the sum of $Q$ in all test cases does not exceed $5 \cdot 10^5$

-----Subtasks-----
Subtask #1 (30 points):
- $1 \le N, Q \le 1,000$
- $v_1 < v_2 < \ldots < v_N$
- the sum of $N$ in all test cases does not exceed $5,000$
- the sum of $Q$ in all test cases does not exceed $5,000$
Subtask #2 (70 points): original constraints

-----Example Input-----
2
4 2
4 2 5 7
2 3
3 4
2 1
1 1
2 1

-----Example Output-----
4 3
3 2
-1 2

-----Explanation-----
Example case 1: For the first query, there are two paths with cost $4$ from city $2$ to city $3$:
- $2 \rightarrow 1 \rightarrow 3$: cost $(|4-2|+1-2)+(|5-4|+3-1) = 4$, length $3$
- $2 \rightarrow 3$: cost $|5-2|+3-2 = 4$, length $2$
All other paths have greater costs, so the minimum cost is $4$. Among these two paths, we want the one with greater length, which is $3$.

```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 = 't = int(input())\nfor i in range(t):\n n, q = input().split()\n cities = [int(x) for x in input().split()]\n cities_sort = cities.copy()\n cities_sort.sort()\n for j in range(int(q)):\n  q1, q2 = [int(x) for x in input().split()]\n  a = cities[q1 - 1]\n  b = cities[q2 - 1]\n  if (a > b):\n   c = a\n   a = b\n   b = c\n  pos_a = cities_sort[::1].index(a)\n  pos_b = len(cities_sort) - 1 - cities_sort[::-1].index(b)\n  print(q2 - q1 + b - a, pos_b - pos_a + 1)'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13419