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 MN 500000
#define mod 1004535809
#define mod2 1000000009
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
char s[MN+5];
int n,b[MN+5],pw[MN+5],pw2[MN+5],ha[MN+5],ha2[MN+5],ans;
inline int gh(int l,int r){return (ha[r]-1LL*ha[l-1]*pw[r-l+1]%mod+mod)%mod;}
inline int GH(int l,int r){return (ha2[r]-1LL*ha2[l-1]*pw2[r-l+1]%mod2+mod2)%mod2;}
bool Match(int l,int r,int L,int R){return gh(l,r)==gh(L,R)&&GH(l,r)==GH(L,R);}
int main()
{
    scanf("%s",s+1);n=strlen(s+1);pw[0]=pw2[0]=1;
    for(int i=1;i<=n;++i) 
        pw[i]=31LL*pw[i-1]%mod,pw2[i]=37LL*pw2[i-1]%mod2,
        ha[i]=(31LL*ha[i-1]+s[i])%mod,ha2[i]=(37LL*ha2[i-1]+s[i])%mod2;
    int all=1;
    for(int i=2;i<=n;++i) all&=(s[i]==s[i-1]);
    if(all) return 0*printf("%d\n1",n);
    for(int i=1;i<=n;++i)
        for(int j=i<<1;j<=n;j+=i)
            if(Match(j-2*i+1,j-i,j-i+1,j)) b[j]=1;
            else break;
    for(int i=1;i<=n;++i)
        for(int j=n-(i<<1)+1;j>=1;j-=i)
            if(Match(j,j+i-1,j+i,j+2*i-1)) b[j-1]=1;
            else break;
    if(!b[n]) return 0*puts("1\n1");
    for(int i=1;i<n;++i) if(!b[i]) ++ans;
    printf("2\n%d",ans);
	return 0;
}
