You are given two strings $S$ and $R$. Each of these strings has length $N$. We want to make $S$ equal to $R$ by performing the following operation some number of times (possibly zero):
- Choose two integers $a$ and $b$ such that $1 \le a \le b \le N$.
- For each $i$ such that $a \le i \le b$, replace the $i$-th character of $S$ by the $i$-th character of $R$.
Suppose that we make $S$ equal to $R$ by performing this operation $k$ times, in such a way that the total number of replaced characters (i.e. the sum of all $k$ values of $b-a+1$) is $l$. Then, the cost of this process is defined as $k \cdot l$.
Find the minimum cost with which we can make $S$ equal to $R$.

-----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 a single string $S$.
- The second line contains a single string $R$.

-----Output-----
For each test case, print a single line containing one integer ― the minimum cost.

-----Constraints-----
- $1 \le T \le 4,000$
- $1 \le N \le 10^6$
- $|S| = |R| = N$
- $S$ and $R$ contain only lowercase English letters
- the sum of $N$ over all test cases does not exceed $2 \cdot 10^6$

-----Example Input-----
1
adefb
bdefa

-----Example Output-----
4

-----Explanation-----
Example case 1: $S$ can be made equal to $R$ in two moves. First, we replace $S_1$ by $R_1$ and then replace $S_5$ by $R_5$. We have $k = l = 2$, so the cost is $2 \cdot 2 = 4$. If we wanted to perform only one operation, the cost would be $5$.

```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 = "# P.S.: I understood the solution using the code of https://www.codechef.com\xa0\xa0\xa0\xa0/viewsolution/33283413 \n# the user's name is kush_07(thanks to him), but it took me some time to figure \xa0\xa0\xa0\xa0it out so i\n# have written some comments which might help you if you by any chance arrive at \xa0\xa0\xa0\xa0the solution.\nfor _ in range(int(input())):\n s=list(input())\n r=list(input())\n l=[]\n flag=0\n eq=0\n k=0\n s_1=0\n for i in range(len(s)):\n  if s[i]!=r[i]:\n   # When the group/array is not the first array to have different \xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0characters \n   # this is identified using the flag variable\n   # as we don't need the elements which were before the first \xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0different\n   # subarray but only need the elements between the different \xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0subarrays after\n   # the first subarray having different characters (subarrays here \xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0refer to 'islands' \n   # mentioned in the tutorial)\n   if eq>0 and flag==1:\n    l.append(eq)\n    k+=1\n   # we are not appending the equal elements before this as this is the \xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0first subarray with\n   # different characters (flag==0) but we still need to count it as \xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0one operation so k+=1\n   elif flag==0:\n    k+=1\n   eq=0\n   flag=1\n   # Number of characters that would require replacing\n   s_1+=1\n  else:\n   # Counting the same characters \n   eq+=1\n l.sort()\n res=0\n res=k*s_1\n for i in range(len(l)):\n  s_1=s_1+l[i]\n  k=k-1\n  res=min(res,s_1*k)\n print(res)\n   \n   \n"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13285