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 <cstdio>
#include <cstring>
#include <algorithm>
#define R register
using namespace std;
const int MAXN=500100;
int next1[MAXN],next2[MAXN];
char s1[MAXN],s2[MAXN];

int reads(char *st)
{
	int len=0;char ch=getchar();
	while(ch<'a'||ch>'z')ch=getchar();
	while(ch>='a'&&ch<='z')st[++len]=ch,ch=getchar();
	return len;
}

void getnext(char *st,int *next,int len)
{
	int j=0;
	for(int i=2;i<=len;++i)
	{
		while(j&&st[i]!=st[j+1])j=next[j];
		if(st[i]==st[j+1])++j;
		next[i]=j;
	}
}

int main()
{
	int n=reads(s1);
	for(R int i=1;i<=n;++i)s2[i]=s1[n-i+1];
	getnext(s1,next1,n);
	getnext(s2,next2,n);
	int t=n-next1[n];
	if(t==1){printf("%d\n1\n",n);}
	else if(t<n&&n%t==0)
	{
		puts("2");
		int ans=0;
		for(int i=1;i<n;++i)
		{
			int t1=i-next1[i],t2=n-i-next2[n-i];
			if(t1<i&&i%t1==0)continue;
			if(t2<n-i&&(n-i)%t2==0)continue;
			++ans;
		}
		printf("%d\n",ans);
	}
	else puts("1\n1");
	return 0;
}