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 rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define all(c) c.begin(),c.end()
#define pb push_back
#define fs first
#define sc second
#define show(x) cout << #x << " = " << x << endl
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
using namespace std;
template<class S,class T> ostream& operator<<(ostream& o,const pair<S,T> &p){return o<<"("<<p.fs<<","<<p.sc<<")";}
template<class T> ostream& operator<<(ostream& o,const vector<T> &vc){o<<"sz = "<<vc.size()<<endl<<"[";for(const T& v:vc) o<<v<<",";o<<"]";return o;}

string s;
int A[500001],B[500001];
void kmp(string s,int* A){
	A[0]=-1;
	int j=-1;
	rep(i,s.size()){
		while(j>=0&&s[i]!=s[j]) j=A[j];
		j++;
		A[i+1]=j;
	}
}
int main(){
	string s;
	cin>>s;
	int N=s.size();
	vector<int> ds;
	rep1(i,N-1){
		if(N%i==0) ds.pb(i);
	}
	int T=-1;
	for(int d:ds){
		bool is=1;
		rep(i,N) if(s[i]!=s[i%d]) is=0;
		if(is){
			T=d;
			break;
		}
	}
	if(T==-1){
		cout<<1<<endl<<1<<endl;
	}else if(T==1){
		cout<<N<<endl<<1<<endl;
	}else{
		kmp(s,A);
		rep(i,N) A[i]=i+1-A[i+1];
		reverse(all(s));
		kmp(s,B);
		rep(i,N) B[i]=i+1-B[i+1];
		int cnt=0;
		rep1(i,N-1){
			if( (A[i-1]==i || i%A[i-1]>0) && (B[N-1-i]==N-i || (N-i)%B[N-1-i]>0) ) cnt++;
		}
		cout<<2<<endl<<cnt<<endl;
	}
}
