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 <bits/stdc++.h>

using namespace std;
const int maxn=500005;
char str[maxn];
char str1[maxn];
int Next[maxn];
void getNext()
{
   Next[0]=-1;
   int k=-1,j=0;
   while(str[j]!='\0')
   {
        while(k!=-1&&str[k]!= str[j])
            k=Next[k];
        ++j;++k;
        Next[j]=k;
   }
   int l=strlen(str);
//   for(int i=0;i<=l;i++)
//    printf("%d ",Next[i]);
//   printf("\n");
}

int main()
{
    scanf("%s",str);
    getNext();
    int l=strlen(str);
    int s=l-Next[l];
    //printf("%d %d\n",l,s);
    if(s==l||l%s!=0){
        printf("1\n1\n");return 0;
    }
    if(s==1){
        printf("%d\n1\n",l);return 0;
    }
    int ans=l-1-(l/s-1)+((l/s)==2);
    //printf("%d ans\n",ans);
    str[s]=0;
    getNext();
    for(int i=1;i<=s;i++)
    {
        int t1=i-Next[i];
        if(t1==i||i%t1!=0){
            continue;
        }
        ans--;
    }

    //printf("%d ans\n",ans);
    for(int i=0;i<s/2;i++)
    {
        swap(str[i],str[s-1-i]);
    }
   // printf("%s\n",str);
    getNext();
    for(int i=1;i<=s;i++)
    {
        int t1=i-Next[i];
        if(t1==i||i%t1!=0){
            continue;
        }
        ans--;
        //printf("%d %d\n",i,Next[i]);
    }
    printf("2\n%d\n",ans);
    return 0;
}