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<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define rep(i, a, b) for(int i=(a); i<(b); i++)
#define sz(a) (int)a.size()
#define de(a) cout<<#a<<" = "<<a<<endl
#define dd(a) cout<<#a<<" = "<<a<<" "
typedef long long ll;
typedef double db;
typedef pair<int, int> pii;
typedef vector<int> vi;
//----
const int N=505050;
int nx0[N], nx1[N];
char s[N];
void kmp(char s[], int nx[]){
	nx[0]=-1;
	for(int i=1, j=-1; s[i]; i++){
		while(j!=-1 && s[j+1]!=s[i]) j=nx[j];
		if(s[j+1]==s[i])j++;
		nx[i]=j;
	}
}
bool check(int n, int nx[]){
	return (nx[n-1]!=-1&&(n%(n-1-nx[n-1])==0));
}
int main(){
	scanf("%s", s);
	int n=strlen(s);
	kmp(s, nx0);
	reverse(s, s+n);
	kmp(s, nx1);
	if(!check(n, nx0))printf("1\n1\n");
	else if(nx0[n-1]==n-2)printf("%d\n1\n", n);
	else{
		int ans=0;
		rep(i, 1, n+1)if(!check(i, nx0)&&!check(n-i, nx1))ans++;
		printf("2\n%d\n", ans);
	}
	return 0;
}
/*
aab
1
1

bcbc
2
3

ddd
3
1
*/