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
// ===================================
//   author: M_sea
//   website: http://m-sea-blog.com/
// ===================================
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#define re register
using namespace std;

inline int read() {
    int X=0,w=1; char c=getchar();
    while (c<'0'||c>'9') { if (c=='-') w=-1; c=getchar(); }
    while (c>='0'&&c<='9') X=X*10+c-'0',c=getchar();
    return X*w;
}

const int N=500000+10;

int n; char s[N];
int prenxt[N],sufnxt[N];

inline void getnext() {
    for (re int i=2,j=prenxt[1]=0;i<=n;++i) {
        while (j&&s[j+1]!=s[i]) j=prenxt[j];
        if (s[j+1]==s[i]) ++j;
        prenxt[i]=j;
    }
    reverse(s+1,s+n+1);
    for (re int i=2,j=sufnxt[n]=0;i<=n;++i) {
        while (j&&s[j+1]!=s[i]) j=sufnxt[j];
        if (s[j+1]==s[i]) ++j;
        sufnxt[i]=j;
    }
}

inline int checkpre(int i) { return !prenxt[i]||i%(i-prenxt[i]); }
inline int checksuf(int i) { return !sufnxt[i]||i%(i-sufnxt[i]); }

int main() {
    scanf("%s",s+1); n=strlen(s+1); getnext();
    if (checkpre(n)) { puts("1\n1"); return 0; }
    if (n-prenxt[n]==1) { printf("%d\n1\n",n); return 0; }
    int ans=0;
    for (re int i=1;i<=n;++i)
        if (checkpre(i)&&checksuf(n-i)) ++ans;
    printf("2\n%d\n",ans);
    return 0;
}
//qwq