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: Wrong Answer)

//              +-- -- --++-- +-In the name of ALLAH-+ --++-- -- --+              \\

#include <bits/stdc++.h>

#define F first
#define S second
#define _sz(x) (int)x.size()

using namespace std ;
using ll = long long ;
using ld = long double ;
using pii = pair <int , int> ;

const int N = 5e5 + 20 , base = 791 ;
int n ;
ll h[N] , pw[N] ;
bool bad[N] ;
string s ;

inline ll get (int l , int r) {
	return h[r] - h[l] * pw[r - l] ;	
}

inline void check_n () {
	bool ok = 1 ;

	for (int i = 1 ; i < n ; i ++) ok &= (s[i] == s[0]) ;

	if (ok) {
		cout << n << "\n1\n" ;
		exit(0) ;
	}
}

inline void check_1 () {
	for (int i = 1 ; i < n ; i ++) {
		if (n % (n - i)) continue ;
		if (get(0 , i) == get(n - i , n)) return ;
	}

	cout << "1\n1\n" ;
	exit(0) ;
}

int main(){
	ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0) ;

	for (int i = pw[0] = 1 ; i < N ; i ++) pw[i] = pw[i - 1] * base ;
	
	cin >> s ;

	n = _sz(s) ;

	for (int i = 0 ; i < n ; i ++) h[i + 1] = base * h[i] + s[i] ;
	
	check_n() ;
	check_1() ;

	bad[n - 1] = 1 ;

	for (int d = 1 ; d <= n ; d ++) {
		for (int a = 2 ; a * d <= n ; a ++) {
			if (get((a - 1) * d , a * d) != get(0 , d)) break ;
			bad[a * d - 1] = 1 ;
		}
	
		for (int a = 2 ; a * d <= n ; a ++) {
			if (get(n - a * d , n - (a - 1) * d) != get(n - d , n)) break ;
			bad[n - a * d - 1] = 1 ;
		}
	}

	cout << "2\n" ;

	int ans = 0 ;
	for (int i = 0 ; i < n ; i ++) ans += !bad[i] ;

	cout << ans << '\n' ;
}