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<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<queue>
using namespace std;

typedef long long ll;
typedef pair<int,int> P;

#define pb push_back
#define fr first
#define sc second

struct MP{
	string s;
	int a[500010];
	void calc_MP(){
		a[0] = -1;
		int j = -1;
		for(int i = 0 ; i < s.size() ; i ++){
			while(j >= 0 && s[i] != s[j])j = a[j];
			j ++;
			a[i+1] = j;
		}
	}
}S,T;

int main(){
	string s;
	cin >> s;
	
	S.s = s;
	T.s = s;
	reverse(T.s.begin(),T.s.end());
	S.calc_MP();
	T.calc_MP();
	
	if(S.a[S.s.size()] == S.s.size()-1){
		printf("%d\n%d\n",(int)s.size(),1);
		return 0;
	}
	if(S.a[S.s.size()] == 0 || S.s.size()%(S.s.size()-S.a[S.s.size()]) != 0){
		printf("%d\n%d\n",1,1);
		return 0;
	}
	int cnt = 0;
	for(int i = 1 ; i <= -1+(int)s.size() ; i ++){
		int j = s.size()-i;
		if((S.a[i] == 0 || i%(i-S.a[i]) != 0)&&(T.a[j] == 0 || j%(j-T.a[j]) != 0))cnt ++;
	}
	printf("%d\n%d\n",2,cnt);
}

