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 N=500010,mod=19260817,qs=19491001;
#define ull unsigned long long
char c[N];
int n;
unsigned long long h[N],qp[N];
unsigned int h1[N],qp1[N];
ull gh(int l,int r)
{
	return (h[r]-1ll*h[l-1]*qp[r-l+1]%qs+qs)%qs;
}
unsigned int gh1(int l,int r)
{
	return (h1[r]-1ll*h1[l-1]*qp1[r-l+1]%mod+mod)%mod;
}
int len[N],len1[N];
int main()
{
	scanf("%s",c+1);
	n=strlen(c+1);
	qp[0]=1;
	for(int i=1;i<=n;i++)
		h[i]=1ll*h[i-1]*mod%qs+c[i]-'a'+1,h1[i]=1ll*h1[i-1]*qs%mod+c[i]-'a'+1;
	qp1[0]=1;
	for(int i=1;i<=n;i++)
		qp[i]=1ll*qp[i-1]*mod%qs,qp1[i]=1ll*qp1[i-1]*qs%mod;
	for(int i=1;i<=n/2;i++)
		if(!len[i*2] && gh(1,i)==gh(i+1,i*2) && gh1(1,i)==gh1(i+1,i*2))
		{
			len[i*2]=i;
			ull p=gh(1,i);
			unsigned int q=gh1(1,i);
			for(int j=i*3;j<=n;j+=i)
				if(gh(j-i+1,j)==p && gh1(j-i+1,j)==q)
					len[j]=i;
				else
					break;
		}
	for(int i=1;i<=n/2;i++)
		if(!len1[n-i*2+1] && gh(n-i+1,n)==gh(n-2*i+1,n-i) && gh1(n-i+1,n)==gh1(n-2*i+1,n-i))
		{
			len1[n-i*2+1]=i;
			ull p=gh(n-i+1,n);
			unsigned int q=gh1(n-i+1,n);
			for(int j=i*3;j<=n;j+=i)
				if(gh(n-j+1,n-j+i)==p && gh1(n-j+1,n-j+i)==q)
					len1[n-j+1]=i;
				else
					break;
		}
	if(len[n]==1)
	{
		printf("%d\n1\n",n);
		return 0;
	}
	if(!len[n])
	{
		printf("1\n1\n");
		return 0;
	}
	int ans=0;
	for(int i=1;i<n;i++)
		if(!len[i] && !len1[i+1])
			ans++;
	printf("2\n%d\n",ans);
	return 0;
}
