Find the digit occurring the most in a range of prime numbers

Question

Find the digit which occurs the most in a range of prime numbers.

Input:

Two numbers, p and q, specifying the range; the range includes both p and q.

Output:

The digit that occurs most frequently in that range of prime numbers. If two or more digits are tied, all of them should be outputted.

Winning Criteria:

Fastest code to complete (insert 1+ specific test cases here) wins.

Example

Input: 21 40

Output: 3

Explanation

The prime numbers in the range 21 40 are 23, 29, 31, and 37. Within these numbers, 2 occurs 2 times, 3 occurs 3 times, and 1, 7 and 9 occur 1 time each. Thus, as 3 appears the most, the correct output is 3.

Answer 1

Python 3

import collections
p=lambda n:n>1 and all(n%i for i in range(int(n**0.5),1,-1))
print(collections.Counter(''.join(map(str,filter(p,range(*map(int,raw_input().split(' '))))))).most_common(1)[0][0])

Probably not the best way it can be done, but a start. (196 bytes)

Time Specs

Using Python's builtin code timer module, timeit, I got these times for each listed range.

(1,       11):      0.000084381103520
(1,      101):      0.000338881015778
(1,     1001):      0.003911740779880
(1,    10001):      0.071979818344100
(1,   100001):      1.785329985620000
(1,  1000001):     54.977741956700000
(1, 10000001):   1700.231099130000000

It can go over a range of a million numbers in under a minute.

Usage

$ python test.py
21 41
3
$ python test.py
51 89
7
$ python test.py
201 501
3

Ungolfed

import collections

def is_it_prime(n):
    if n > 1:
        if all(n % i for i in range(int(n ** 0.5), 1, -1)):
            return True
    return False

start, stop = raw_input.split(' ', 1)

primes = ""
for p in range(int(start), int(stop)):
    if is_it_prime(p):
        primes += str(p)

counter = collections.Counter(primes).most_common(1)
print(counter[0][0])