Problem:
Score : 900 points

Problem Statement

Let x be a string of length at least 1.
We will call x a good string, if for any string y and any integer k (k \geq 2), the string obtained by concatenating k copies of y is different from x.
For example, a, bbc and cdcdc are good strings, while aa, bbbb and cdcdcd are not.

Let w be a string of length at least 1.
For a sequence F=(\,f_1,\,f_2,\,...,\,f_m) consisting of m elements,
we will call F a good representation of w, if the following conditions are both satisfied:

For any i \, (1 \leq i \leq m), f_i is a good string.

The string obtained by concatenating f_1,\,f_2,\,...,\,f_m in this order, is w.

For example, when w=aabb, there are five good representations of w:

(aabb)

(a,abb)

(aab,b)

(a,ab,b)

(a,a,b,b)

Among the good representations of w, the ones with the smallest number of elements are called the best representations of w.
For example, there are only one best representation of w=aabb: (aabb).

You are given a string w. Find the following:

the number of elements of a best representation of w

the number of the best representations of w, modulo 1000000007 \, (=10^9+7)

(It is guaranteed that a good representation of w always exists.)

Constraints

1 \leq |w| \leq 500000 \, (=5 \times 10^5)

w consists of lowercase letters (a-z).

Partial Score

400 points will be awarded for passing the test set satisfying 1 \leq |w| \leq 4000.

Input

The input is given from Standard Input in the following format:

w

Output

Print 2 lines.

In the first line, print the number of elements of a best representation of w.

In the second line, print the number of the best representations of w, modulo 1000000007.

Sample Input 1

aab

Sample Output 1

1
1

Sample Input 2

bcbc

Sample Output 2

2
3

In this case, there are 3 best representations of 2 elements:

(b,cbc)

(bc,bc)

(bcb,c)

Sample Input 3

ddd

Sample Output 3

3
1

Solution in Ruby (2.3.3) (judge verdict: Accepted)

def z_algorithm(s)
    c = 0
    ret = Array.new(N, 0)
    (1...N).each do |i|
        if i + ret[i-c] < c + ret[c]
            ret[i] = ret[i-c]
        else
            j = c + ret[c] - i
            j = 0 if j < 0
            j += 1 while i+j < N && s[j] == s[i+j]
            ret[i] = j
            c = i
        end
    end
    ret[0] = N
    ret
end

S = gets.chomp
N = S.length
len = [z_algorithm(S)]
S.reverse!
len[1] = z_algorithm(S)
bad = Array.new(2){Array.new(N+1)}

min_period = N
2.times do |k|
    (1...N).each do |period|
        next if bad[k][period]
        0.step(N-1, period) do |s|
            if len[k][s] >= period
                bad[k][s+period] = true if s > 0
            else
                break
            end
        end
        min_period = period if min_period == N && bad[k][N]
    end
end

if min_period == N
    opt = 1
    num = 1
elsif min_period == 1
    opt = N
    num = 1
else
    opt = 2
    num = (1...N).count{|i| !bad[0][i] && !bad[1][N-i]}
end
puts opt, num