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>
#define ld double
#define ull unsigned long long
#define ll long long
#define pii pair<int,int >
#define iiii pair<int,pii >
#define mp make_pair
#define INF 1000000000
#define MOD 1000000007
#define rep(i,x) for(int (i)=0;(i)<(x);(i)++)
inline int getint(){
    int x=0,p=1;char c=getchar();
    while (c<=32)c=getchar();
    if(c==45)p=-p,c=getchar();
    while (c>32)x=x*10+c-48,c=getchar();
    return x*p;
}
using namespace std;
//ruogu
const int N=5e5+10;
int n,pre[2][N];
char s[N];
//
void getfail(int *fail){
	fail[1]=0;
	for(int i=2,j=0;i<=n;i++){
		while(j>0&&s[j+1]!=s[i])j=fail[j];
		if(s[j+1]==s[i])j++;
		fail[i]=j;
	}
}
int main(){
	scanf("%s",s+1);
	n=strlen(s+1);
	getfail(pre[0]);
	reverse(s+1,s+1+n);
	getfail(pre[1]);
	int len=n-pre[0][n];
	if(n%len!=0||len==n){
		printf("%d\n%d\n",1,1);
		return 0;
	}
	int c=n/len;
	if(c==n){
		printf("%d\n%d\n",n,1);
		return 0;
	}
	reverse(s+1,s+1+n);
	int res=0;
	for(int i=1;i<n;i++){
		int x=i-pre[0][i];
		if(i%x==0&&x<i)continue;
		 x=(n-i)-pre[1][n-i];
		if((n-i)%x==0&&x<n-i)continue;
		res++;
	}
	printf("%d\n%d\n",2,res);
	return 0;
}