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>
using namespace std;
char s[1000001];int z[1500001],n,nxt[1000001],buc[1000011],rnxt[1000001];bool cut[1000001];
void zfunc(char *s,int n,int *z)
{
	z[1]=n;int l=0,r=0,pt;
	for(int i=2;i<=n;++i)
	{
		if(r<i){for(pt=i;pt<=n&&s[pt]==s[pt-i+1];++pt);--pt;z[i]=pt-i+1;if(z[i])l=i,r=pt;}
		else{int b=r-i+1,i_=i-l+1;if(z[i_]<b)z[i]=z[i_];else{for(pt=r+1;pt<=n&&s[pt]==s[pt-i+1];++pt);--pt;z[i]=pt-i+1;l=i;r=pt;}}
	}
}
void cal_next(int *z,int n,int *nxt)
{
	memset(buc,0x3f,sizeof(int)*(2*n+2));
	for(int i=n;i>=2;--i)buc[z[i]+i]=i;
	for(int i=2*n;i;--i)buc[i]=min(buc[i],buc[i+1]);
	for(int i=2;i<=n;++i)if(buc[i+1]<=i)nxt[i]=i-buc[i+1]+1;
}
int main()
{
	scanf("%s",s+1);n=strlen(s+1);int len=0;if(n==1){printf("1\n1\n");return 0;}zfunc(s,n,z);cal_next(z,n,nxt);
	len=n-nxt[n];if(!nxt[n]||n%len){printf("1\n1\n");return 0;}if(len==1){printf("%d\n%d\n",n,1);return 0;}
	printf("2\n");int ans=n-1;reverse(s+1,s+1+n);zfunc(s,n,z);cal_next(z,n,rnxt);
	for(int i=2;i<n;++i)if(nxt[i]&&!(i%(i-nxt[i])))cut[i]=true;
	for(int i=2;i<n;++i)if(rnxt[i]&&!(i%(i-rnxt[i])))cut[n-i]=true;
	for(int i=1;i<n;++i)ans-=cut[i];printf("%d\n",ans);return 0;
}