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 <cstdio>
#include <cstring>

using namespace std;
const int MAXN = 5e5 + 7;

inline int R()
{
	int rt = 0;
	bool isn = false;
	char ch = getchar();
	for ( ; ch < '0' || ch > '9'; ch = getchar() )
		isn = ch == '-' ? true : isn;
	for ( ; ch >= '0' && ch <= '9'; ch = getchar() )
		rt = rt * 10 + ch - '0';
	return isn ? -rt : rt;
}

char s[MAXN], r[MAXN];
int dp[MAXN], rdp[MAXN], n;

int Gao ()
{
	dp[1] = 0;
	for ( int i = 2; i <= n; ++i )
	{
		int nc = dp[i - 1];
		while ( nc && s[nc + 1] != s[i] )
			nc = dp[nc];
		dp[i] = s[nc + 1] == s[i] ? nc + 1 : 0;
	}
	if ( dp[n] == n - 1 )
		return n;
	if ( !dp[n] || n % ( n - dp[n] ) )
		return 1;
	return 2;
}

void Rn()
{
	for ( int i = 1; i <= n; ++i )
		r[i] = s[n - i + 1];
	rdp[1] = 0;
	for ( int i = 2; i <= n; ++i )
	{
		int nc = rdp[i - 1];
		while ( nc && r[nc + 1] != r[i] )
			nc = rdp[nc];
		rdp[i] = r[nc + 1] == r[i] ? nc + 1 : 0;
	}
}

int main()
{
//	freopen ( "test.in", "r", stdin );
	scanf ( "%s", s + 1 );
	n = strlen ( s + 1 );
	int ans = Gao();
	printf ( "%d\n", ans );
	if ( ans == n || ans == 1 )
		puts ( "1" );
	else
	{
		Rn();
		int cnt = 0;
		for ( int i = 1; i < n; ++i )
			if ( ( !dp[i] || i % ( i - dp[i] ) ) && ( !rdp[n - i] || ( n - i ) % ( n - i - rdp[n - i] ) ) )
				++cnt;
		printf ( "%d\n", cnt );
	}
	return 0;
}