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<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<map>
#include<stack>
using namespace std;
typedef long long ll;
#define N 500015
const int bas=23333,bs=2333;
const int p=1e9+7,pp=1e9+9;
char ch[N];
int n;
int ha[N],bin[N],hh[N],bn[N];
vector<int>v[N];
inline int geh(int l,int r){return (ha[r]-1ll*ha[l-1]*bin[r-l+1]%p+p)%p;}
inline int gh(int l,int r){return (hh[r]-1ll*hh[l-1]*bn[r-l+1]%pp+pp)%pp;}
inline bool jud(int l,int r)
{
	for(int i=0;i<v[r-l+1].size();i++)
	{
		int x=v[r-l+1][i];
		if(geh(l,r-x)==geh(l+x,r)&&gh(l,r-x)==gh(l+x,r))return true;
	}return false;
}
int main()
{
    scanf("%s",ch+1);n=strlen(ch+1);bool flag=true;
    for(int i=2;i<=n;i++)if(ch[i]!=ch[1]){flag=false;break;}
    if(flag){printf("%d\n%d\n",n,1);return 0;}
    for(int i=1;i<=n;i++)for(int j=i+i;j<=n;j+=i)v[j].push_back(i);
    bin[0]=bn[0]=1;for(int i=1;i<=n;i++)bin[i]=1ll*bin[i-1]*bas%p,bn[i]=1ll*bn[i-1]*bs%pp;
    for(int i=1;i<=n;i++)ha[i]=(1ll*ha[i-1]*bas+ch[i]-'a'+1)%p,hh[i]=(1ll*hh[i-1]*bs+ch[i]-'a'+1)%pp;
    if(!jud(1,n)){puts("1");puts("1");return 0;}
	puts("2");
	int ans=0;
	for(int i=1;i<n;i++)
	{
		if((!jud(1,i))&&(!jud(i+1,n)))ans++;
	}printf("%d\n",ans);
}
