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)

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define SF scanf
#define PF printf
#define MAXN 500010
using namespace std;
typedef unsigned long long ull;
ull pre[MAXN],las[MAXN],fsp[MAXN];
int prel[MAXN],lasl[MAXN];
char s[MAXN];
int n,minlen;
void check(int x,int flag){
	if(flag==0){
		prel[x]=x;
		for(int i=1;i*i<=x;i++)
			if(x%i==0){
				if(pre[x]-pre[i]*fsp[x-i]==pre[x-i]){
					prel[x]=i;
					break;
				}
				if(pre[x]-pre[x/i]*fsp[x-x/i]==pre[x-x/i]){
					prel[x]=x/i;	
				}
			}
	}
	else{
		lasl[x]=x;
		for(int i=1;i*i<=x;i++)
			if(x%i==0){
				if(las[x]-las[i]*fsp[x-i]==las[x-i]){
					lasl[x]=i;	
					break;
				}
				if(las[x]-las[x/i]*fsp[x-x/i]==las[x-x/i]){
					lasl[x]=x/i;	
				}
			}
	}
}
int main(){
	SF("%s",s+1);
	n=strlen(s+1);
	fsp[0]=1;
	for(int i=1;i<=n;i++)
		fsp[i]=fsp[i-1]*131ull;
	for(int i=1;i<=n;i++)
		pre[i]=pre[i-1]*131ull+(ull)(s[i]-'a'+1);
	reverse(s+1,s+n+1);
	las[0]=s[0]-'a'+1;
	for(int i=1;i<=n;i++)
		las[i]=las[i-1]*131ull+(ull)(s[i]-'a'+1);
	for(int i=1;i<=n;i++){
		check(i,0);
		check(i,1);
	}
	if(prel[n]==1){
		PF("%d\n1",n);
		return 0;	
	}
	if(prel[n]==n){
		PF("1\n1");
		return 0;
	}
	int ans=0;
	
	for(int i=1;i<n;i++){
		if(prel[i]==i&&lasl[n-i]==n-i)
			ans++;
	}
	PF("2\n%d",ans);
}