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>
const int Maxn = 500001;
#define For( i, j, k ) for( int i = j; i <= k; ++i ) 
using std::cin;
using std::cout;
char c[ Maxn ];
int n, nxt1[ Maxn ], nxt2[ Maxn ];
void KMP( int *nxt )
{
	nxt[ 0 ] = -1;
	For( i, 1, n )
	{
		int j = nxt[ i - 1 ];
		while( j > -1 && c[ i ] != c[ j + 1 ] )
			j = nxt[ j ];
		nxt[ i ] = j + 1;
	}
	return;
}

bool chkPre( int p ) { return nxt1[ p ] && p % ( p - nxt1[ p ] ) == 0; }
bool chkSuf( int p ) { return nxt2[ p ] && p % ( p - nxt2[ p ] ) == 0; }
bool chk( int p ) { return !chkPre( p ) && !chkSuf( n - p ); }
int main( )
{
	std::ios::sync_with_stdio( false );
	cin >> c + 1;
	n = strlen( c + 1 );
	KMP( nxt1 );
	std::reverse( c + 1, c + n + 1 );
	KMP( nxt2 );
	if( nxt1[ n ] && !( n % ( n - nxt1[ n ] ) ) )
	{
		if( nxt1[ n ] == n - 1 )
			cout << n << '\n' << 1;
		else
		{
			cout << 2 << '\n';
			int ans = 0;
			For( i, 1, n )
				ans += chk( i );
			cout << ans;
		}
	}
	else
		cout << 1 << '\n' << 1;
	return 0;
}