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 <stdio.h>
#include <iostream>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <string>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <ctime>
#include <vector>
#include <fstream>
#include <list>
#include <iomanip>
#include <numeric>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define ms(s) memset(s, 0, sizeof(s))
const int INF = 1e9;
inline int read(){
	int X = 0,w = 0 ;
	char ch = 0;
	while(!isdigit(ch)) {w |= ch == '-';ch = getchar();}
	while(isdigit(ch)) X = (X<<3) + (X<<1) + (ch ^ 48),ch = getchar();
	return w ? -X : X;
}
int nxt[1000005];
int nxt2[1000005];
int n;
char s[1000005];
void get_nxt()
{
	nxt[0]=-1;
	for(int i=0,j=-1;i<n;i++)
	{
		while(~j and s[i]!=s[j])
		{
			j=nxt[j];
		}
		nxt[i+1]=++j;
	}
}
void get_nxt2()
{
	nxt2[0]=-1;
	for(int i=0,j=-1;i<n;i++)
	{
		while(~j and s[i]!=s[j])
		{
			j=nxt2[j];
		}
		nxt2[i+1]=++j;
	}
}
int main() 
{
	scanf("%s",s);
	n=strlen(s);
	get_nxt();
	reverse(s,s+n);
	get_nxt2();
	if(n%(n-nxt[n])!=0 or nxt[n]==0)
	{
		cout<<1<<endl<<1;
		return 0;
	}
	if(n-nxt[n]==1)
	{
		cout<<n<<endl<<1<<endl;
		return 0;
	}
	int ans=0;
	for(int i=1;i<n;i++)
	{
		int a=i;
		int b=n-i;
		if((a%(a-nxt[a])!=0 or nxt[a]==0) and (b%(b-nxt2[b])!=0 or nxt2[b]==0))
		{
			ans++;
		}
	}
	cout<<2<<endl;
	cout<<ans;
    return 0;
}