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>
#define pb push_back
#define int long long
#define mp make_pair
#define inf 998244353
#define LINF 1000000000000000007LL
#define ll long long
using namespace std;
bool gd[514514],rd[514514];
signed main(){
	string s,r;
	cin>>s;
	r = s;
	reverse( r.begin(), r.end() );
	int n = s.size();
	vector<int> a(n),b(n);
	a[0] = n;
	int i=1,j=0;
	while( i < s.size() ){
		while( i + j < s.size() && s[j] == s[i+j] ) j++;
		a[i] = j;
		if( j == 0 ){ i++; continue; }
		int k=1;
		while( i+k < s.size() && k+a[k] < j ){
			a[i+k] = a[k];
			k++;
		}
		i+=k; j-=k;
	}
	for(int i=0;i<n;i++)gd[i]=true;
	for(int i=1;i<n;i++){
		for(int j=2;(j-1)*i<=a[i];j++){
			gd[i*j-1] = false;

		}
	}
	i=1;j=0;
	while( i < r.size() ){
		while( i + j < r.size() && r[j] == r[i+j] ) j++;
		b[i] = j;
		if( j == 0 ){ i++; continue; }
		int k=1;
		while( i+k < r.size() && k+b[k] < j ){
			b[i+k] = b[k];
			k++;
		}
		i+=k; j-=k;
	}
	for(int i=0;i<n;i++)rd[i]=true;
	for(int i=1;i<n;i++){
		for(int j=2;(j-1)*i<=b[i];j++){
			rd[i*j-1] = false;

		}
	}
	if( gd[n-1] == true ){
		cout<<1<<endl<<1<<endl;
		return 0;
	}
	int cnt = 0;
	for(int i=0;i<n-1;i++){
		if( gd[i] == true && rd[n-2-i] == true)cnt++;
	}
	if( cnt == 0 ){
		cout<<n<<endl<<1<<endl;
		return 0;
	}
	else{
		cout<<2<<endl<<cnt<<endl;
		return 0;
	}
	return 0;
}