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 <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>

using namespace std;

string S;
vector<int> prepi, sufpi;

inline void calcPrefix ( const string S, vector<int>& pi ) {
	int n = S.length ();
	pi.clear (), pi.push_back ( 0 );
	for ( int i = 1; i ^ n; ++ i ) {
		int j = pi[i - 1];
		for ( ; j && S[i] ^ S[j]; j = pi[j - 1] );
		pi.push_back ( j + ( S[i] == S[j] ) );
	}
}

int main () {
	ios :: sync_with_stdio ( false ), cin.tie ( 0 );
	cin >> S;
	int n = S.length ();
	bool allsame = true;
	for ( int i = 1; i ^ S.size () && allsame; ++ i ) allsame &= S[i] == S[0];
	if ( allsame ) return cout << n << "\n1\n", 0;
	calcPrefix ( S, prepi );
	reverse ( S.begin (), S.end () );
	calcPrefix ( S, sufpi );
	if ( ! prepi[n - 1] || n % ( n - prepi[n - 1] ) ) return cout << "1\n1\n", 0;
	int ans = 0;
	for ( int i = 0; i + 1 < n; ++ i ) {
		if ( ( ! prepi[i] || ( i + 1 ) % ( i + 1 - prepi[i] ) )
			&& ( ! sufpi[n - 2 - i] || ( n - i - 1 ) % ( n - i - 1 - sufpi[n - 2 - i] ) ) ) {
			++ ans;
		}
	}
	cout << "2\n" << ans << endl;
	return 0;
}