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`).

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.

Examples

Input

aab


Output

1
1


Input

bcbc


Output

2
3


Input

ddd


Output

3
1
from collections import Counter
*W, = map(ord, input())
N = len(W)

C = Counter(W)
if len(C) == 1:
    print(N)
    print(1)
    exit(0)

def z_algo(S):
    A = [0]*N
    i = 1; j = 0
    A[0] = l = len(S)
    while i < l:
        while i+j < l and S[j] == S[i+j]:
            j += 1
        A[i] = j
        if not j:
            i += 1
            continue
        k = 1
        while l-i > k < j - A[k]:
            A[i+k] = A[k]
            k += 1
        i += k; j -= k
    return A

def calc(W):
    Z = z_algo(W)
    G = [0]*N
    for i in range(N):
        G[i] = 1
    for p in range(1, N):
        if not G[p-1]:
            continue
        for k in range(2, Z[p]//p+2):
            G[k*p-1] = 0
    return G
G0 = calc(W)
W.reverse()
G1 = calc(W)

if G0[N-1]:
    print(1)
    print(1)
    exit(0)

print(2)
print(sum(p and q for p, q in zip(G0[:-1], reversed(G1[:-1]))))