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 <bits/stdc++.h>

using namespace std;
const int MAXN=5e5+10;
char s[MAXN];
int len,res=0,nxt1[MAXN],nxt2[MAXN];

void cal_nxt(int* nxt)
{
	int k=0;
	for(int i=2;i<=len;i++)
	{
		while(k&&s[k+1]!=s[i]) k=nxt[k];
		if(s[k+1]==s[i]) k++;nxt[i]=k;
	}
}

bool check(int* nxt,int pos)
{
	if(!nxt[pos]) return false;
	return (pos%(pos-nxt[pos])==0);
}

int main()
{
	scanf("%s",s+1);len=strlen(s+1);
	cal_nxt(nxt1);
	if(!check(nxt1,len)) printf("1\n1");
	else if(nxt1[len]==len-1)  printf("%d\n1",len);
	else
	{
		reverse(s+1,s+len+1);
		cal_nxt(nxt2);
		for(int i=1;i<=len;i++)
			res+=(!check(nxt1,i))&(!check(nxt2,len-i));
		printf("2\n%d",res);
	}
	return 0;
}