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<iostream>
#include<string>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
const int maxn = 5e5+10;
char w[maxn];
int len;
int pre_next[maxn],suf_next[maxn];
void get_next(){
    pre_next[0]=pre_next[1]=0;
    int j=0;
    for(int i=2;i<=len;i++){
        while(j&&w[i]!=w[j+1])j=pre_next[j];
        if(w[j+1]==w[i])++j;
        pre_next[i]=j;
    }
    reverse(w+1,w+len+1);
    suf_next[0]=suf_next[1]=0;
    j=0;
    for(int i=2;i<=len;i++){
        while(j&&w[i]!=w[j+1])j=suf_next[j];
        if(w[j+1]==w[i])++j;
        suf_next[i]=j;
    }
}
bool pre_check(int i){
    return !pre_next[i]||i%(i-pre_next[i]);
}
bool suf_check(int i){
    return !suf_next[i]||i%(i-suf_next[i]);    
}
int main(){
    scanf("%s",w+1);
    len=strlen(w+1);
    get_next();
    if(pre_check(len)){
        printf("1\n1\n");
    }
    else if(len-pre_next[len]==1){
        printf("%d\n1\n",len);
    }
    else {
        int ans=0;
        for(int i=1;i<=len;i++){
            if(pre_check(i)&&suf_check(len-i))
                ans++;
        }
        printf("2\n%d\n",ans);
    }
    return 0;
}