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<cstdio>
#include<cstring>
#define N 500010
using namespace std;
int len;
char str[N],suf[N];
void make_nxt(int nxt[],char s[])
{
	nxt[0]=-1;
	int f=0;
	for(int i=1;i<len;i++)
	{
		while(f && s[i]!=s[f]) f=nxt[f];
		if(s[i]==s[f]) nxt[i+1]=++f;
		else nxt[i+1]=0;
	}
}
bool no_rep(int nxt[],int p){return nxt[p]==0 || p%(p-nxt[p]);}
int pre[N],nxt[N];
int main()
{
	scanf("%s",str);
	len=strlen(str);
	for(int i=0;i<len;i++) suf[i]=str[len-i-1];
	suf[len]='\0';
	make_nxt(pre,str);
	make_nxt(nxt,suf);
	if(no_rep(pre,len)){puts("1\n1");return 0;}
	bool same=true;
	for(int i=1;i<len;i++) if(str[i]!=str[i-1]) same=false;
	if(same){printf("%d\n1\n",len);return 0;}
	puts("2");
	int ans=0;
	for(int i=0;i<len;i++)
	if(no_rep(pre,i+1) && no_rep(nxt,len-i-1))ans++;
	printf("%d\n",ans);
	return 0;
}
