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 DIM = 5e5 + 5;

bool lef[DIM], rig[DIM];
char str[DIM]; int prf[DIM];

void solve( char str[DIM], bool arr[DIM], int n ) {
    
    for( int i = 2, l = 0; i <= n; i ++ ) {
        while( l != 0 && str[i] != str[l + 1] )
            l = prf[l];
        
        if( str[i] == str[l + 1] )
            l ++;
        
        prf[i] = l;
        arr[i] = ( l != 0 && i % (i - l) == 0 );
    }
    
    return;
}

int main( void ) {
    
    cin >> ( str + 1 );
    int n = (int) strlen( str + 1 );
    
    solve( str, lef, n );
    reverse( str + 1, str + n + 1 );
    
    solve( str, rig, n );
    reverse( rig + 1, rig + n + 1 );
    
    if( lef[n] == false )
        cout << "1\n1\n";
    else
    if( count( str + 1, str + n + 1, str[1] ) == n )
        cout << n << "\n1\n";
    else {
        int ans = 0;
        for( int i = 1; i < n; i ++ )
            ans += ( lef[i] == false && rig[i + 1] == false );
    
        cout << "2\n" << ans << "\n";
    }
    
    return 0;
}