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
#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int MAXN = 5e5 + 10;
char s[MAXN];
int nt[MAXN] , l;
bool f[MAXN] , e[MAXN];
int main(){
	memset(f , 0 , sizeof f);
	cin >> s + 1;
	l = strlen(s + 1);
	bool same = 1;
	for(int i = 2 ; i <= l ; ++ i)
		if(s[i] != s[1]){
			same = 0;
			break;
		}
	if(same){
		cout << l << endl << 1 << endl;
		return 0;
	}
	nt[1] = 0;
	for(int i = 2 ; i <= l ; ++ i){
		int tmp = nt[i - 1];
		while(tmp && s[tmp + 1] != s[i])
			tmp = nt[tmp];
		if(s[tmp + 1] == s[i])
			tmp ++;
		nt[i] = tmp;
		int ll = i - nt[i];
		if(nt[i] && i % ll == 0)
			f[i] = 1;
	}
	nt[l] = 0;
	for(int i = l - 1 ; i >= 1 ; -- i){
		int tmp = nt[i + 1];
		while(tmp && s[l - tmp] != s[i])
			tmp = nt[l - tmp + 1];
		if(s[l - tmp] == s[i])
			tmp ++;
		nt[i] = tmp;
		int ll = l - i + 1 - nt[i];
		if(nt[i] && (l - i + 1) % ll == 0)
			e[i] = 1;
	}
	if(!f[l]){
		cout << 1 << endl << 1 << endl;
		return 0;
	}
	int ans = 0;
	for(int i = 1 ; i < l ; ++ i)
		if(!f[i] && !e[i + 1])
			ans ++;
	cout << 2 << endl << ans << endl;
	return 0;
}