human
Tanya is learning how to add numbers, but so far she is not doing it correctly. She is adding two numbers a and b using the following algorithm:

  1. If one of the numbers is shorter than the other, Tanya adds leading zeros so that the numbers are the same length. 
  2. The numbers are processed from right to left (that is, from the least significant digits to the most significant). 
  3. In the first step, she adds the last digit of a to the last digit of b and writes their sum in the answer. 
  4. At each next step, she performs the same operation on each pair of digits in the same place and writes the result to the left side of the answer. 


For example, the numbers a = 17236 and b = 3465 Tanya adds up as follows:

$$$ \large{ \begin{array}{r} + \begin{array}{r} 17236\\\ 03465\\\ \end{array} \\\ \hline \begin{array}{r} 1106911 \end{array} \end{array}} $$$

  * calculates the sum of 6 + 5 = 11 and writes 11 in the answer. 
  * calculates the sum of 3 + 6 = 9 and writes the result to the left side of the answer to get 911. 
  * calculates the sum of 2 + 4 = 6 and writes the result to the left side of the answer to get 6911. 
  * calculates the sum of 7 + 3 = 10, and writes the result to the left side of the answer to get 106911. 
  * calculates the sum of 1 + 0 = 1 and writes the result to the left side of the answer and get 1106911. 


As a result, she gets 1106911.

You are given two positive integers a and s. Find the number b such that by adding a and b as described above, Tanya will get s. Or determine that no suitable b exists.

Input

The first line of input data contains an integer t (1 ≤ t ≤ 10^4) — the number of test cases.

Each test case consists of a single line containing two positive integers a and s (1 ≤ a < s ≤ 10^{18}) separated by a space.

Output

For each test case print the answer on a separate line.

If the solution exists, print a single positive integer b. The answer must be written without leading zeros. If multiple answers exist, print any of them.

If no suitable number b...
The above is tricky. Write me a correct solution in PYTHON.
assistant
#Here is Python program to find the number b such that by adding a and b as described above, Tanya will get s.

from __future__ import division, print_function

import os
import sys
from io import BytesIO, IOBase

if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip

class LazySegmentTree:
    def __init__(self, data, default=0, func=max):
        """initialize the lazy segment tree with data"""
        self._default = default
        self._func = func

        self._len = len(data)
        self._size = _size = 1 << (self._len - 1).bit_length()
        self._lazy = [pow(2,30)-1] * (2 * _size)

        self.data = [default] * (2 * _size)
        self.data[_size:_size + self._len] = data
        for i in reversed(range(_size)):
            self.data[i] = func(self.data[i + i], self.data[i + i + 1])

    def __len__(self):
        return self._len

    def _push(self, idx):
        """push query on idx to its children"""
        # Let the children know of the queries
        q, self._lazy[idx] = self._lazy[idx], pow(2,30)-1

        self._lazy[2 * idx] &= q
        self._lazy[2 * idx + 1] &= q
        self.data[2 * idx] &= q
        self.data[2 * idx + 1] &= q

    def _update(self, idx):
        """updates the node idx to know of all queries applied to it via its ancestors"""
        for i in reversed(range(1, idx.bit_length())):
            self._push(idx >> i)

    def _build(self, idx):
        """make the changes to idx be known to its ancestors"""
        idx >>= 1
        while idx:
            self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1]) & self._lazy[idx]
            idx >>= 1

    def add(self, start, stop, value):
        """lazily add value to [start, stop)"""
        start = start_copy = start + self._size
        stop = stop_copy = stop + self._size
        while start < stop:
            if start & 1:
                self._lazy[start] &= value
                self.data[start] &= value
                start += 1
            if stop & 1:
                stop -= 1
                self._lazy[stop] &=...