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 <cstring>
#include <algorithm>
using namespace std;

const int N = 5e5 + 10, mod = 1e9 + 7;

char s[N];
int len;
int ne[N], rne[N];
int ans;

void kmp(int *Next)
{
    for (int i = 2, j = 0; i <= len; i ++ )
    {
        while(j && s[j + 1] != s[i]) j = Next[j];
        if(s[j + 1] == s[i]) j ++;
        Next[i] = j;
    }
}

int main()
{
    scanf("%s", s + 1);
    len = strlen(s + 1);
    kmp(ne);
    reverse(s + 1, s + 1 + len);
    kmp(rne);
    // cout << lenS << ' ' << strlen(rs + 1) << endl;

    if(ne[len] && len % (len - ne[len]) == 0)
    {
        if(ne[len] == len - 1) 
        {
            printf("%d\n%d", len, 1);
            return 0;
        }
        puts("2");
        int ans = 0;
        for (int i = 1; i < len; i ++ )
        {
            int k = len - i;
            ans = (ans + (!(ne[i] && i % (i - ne[i]) == 0) && !(rne[k] && k % (k - rne[k]) == 0))) % mod;
        }
        cout << ans << endl;
    }
    else cout << 1 << endl << 1 << endl;
    
    return 0;
}