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 maxn=512345;
int n,nxt[maxn],pre[maxn];
char s[maxn];
int main()
{
    scanf("%s",s+1);
    n=strlen(s+1);
    for (int i=2;i<=n;++i) {
        int j=nxt[i-1];
        while (j&&s[i]!=s[j+1])
            j=nxt[j];
        nxt[i]=j+(s[i]==s[j+1]);
    }
    reverse(s+1,s+1+n);
    for (int i=2;i<=n;++i) {
        int j=pre[i-1];
        while (j&&s[i]!=s[j+1])
            j=pre[j];
        pre[i]=j+(s[i]==s[j+1]);
    }
    if (!nxt[n]||n%(n-nxt[n])) return 0*puts("1\n1");
    int res=0;
    for (int i=1;i<n;++i)
        if ((!nxt[i]||i%(i-nxt[i]))&&(!pre[n-i]||(n-i)%(n-i-pre[n-i]))) ++res;
    printf("%d\n%d\n",n/(n-nxt[n])==n?n:2,max(1,res));
    return 0;
}
