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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <string>
#include <iostream>
using namespace std;
const int b = 257;
const int mod1 = 1000000007;
const int mod2 = 1000000009;
string s; int h1[500009], h2[500009], pw1[500009], pw2[500009]; bool okl[500009], okr[500009];
int hash1(int l, int r) {
	return (h1[r] - 1LL * h1[l] * pw1[r - l] % mod1 + mod1) % mod1;
}
int hash2(int l, int r) {
	return (h2[r] - 1LL * h2[l] * pw2[r - l] % mod2 + mod2) % mod2;
}
int main() {
	cin >> s;
	pw1[0] = pw2[0] = 1;
	for (int i = 1; i <= s.size(); i++) {
		h1[i] = (1LL * h1[i - 1] * b + s[i - 1]) % mod1;
		h2[i] = (1LL * h2[i - 1] * b + s[i - 1]) % mod2;
		pw1[i] = 1LL * pw1[i - 1] * b % mod1;
		pw2[i] = 1LL * pw2[i - 1] * b % mod2;
	}
	int z = -1;
	for (int i = 1; i <= s.size(); i++) {
		if (s.size() % i != 0) continue;
		bool f = true;
		for (int j = i; j < s.size(); j++) {
			if (s[j] != s[j - i]) f = false;
		}
		if (f) {
			z = i;
			break;
		}
	}
	if (z == s.size()) puts("1\n1\n");
	else if (z == 1) printf("%u\n1\n", s.size());
	else {
		for (int i = 1; i <= s.size(); i++) {
			int p1 = hash1(0, i), p2 = hash2(0, i);
			for (int j = i * 2; j <= s.size(); j += i) {
				if (p1 != hash1(j - i, j) || p2 != hash2(j - i, j)) break;
				okl[j] = true;
			}
		}
		for (int i = 1; i <= s.size(); i++) {
			int p1 = hash1(s.size() - i, s.size()), p2 = hash2(s.size() - i, s.size());
			for (int j = i * 2; j <= s.size(); j += i) {
				if (p1 != hash1(s.size() - j, s.size() - j + i) || p2 != hash2(s.size() - j, s.size() - j + i)) break;
				okr[s.size() - j] = true;
			}
		}
		int ret = 0;
		for (int j = 1; j < s.size(); j++) {
			if (!okl[j] && !okr[j]) ret++;
		}
		printf("2\n%d\n", ret);
	}
	return 0;
}