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>
using namespace std;
int n,Z1[500100],Z2[500100],pri[500100],mp[500100],res;
char s[500100];
void Zalgo(int *Z){
	int l=0,r=0;
	for(int i=1;i<n;i++){
		if(i<=r)Z[i]=min(Z[i-l],r-i);
		while(i+Z[i]<n&&s[Z[i]+i]==s[Z[i]])Z[i]++;
		if(i+Z[i]>r)r=i+Z[i],l=i;
	}
}
void ural(){
	for(int i=2;i<=n;i++){
		if(!pri[i])pri[++pri[0]]=i,mp[i]=i;
		for(int j=1;j<=pri[0]&&i*pri[j]<=n;j++){
			pri[i*pri[j]]=true,mp[i*pri[j]]=pri[j];
			if(!(i%pri[j]))break;
		}
	}
}
vector<int>v[500100];
void fact(){
	for(int i=2;i<=n;i++)for(int j=i;j!=1;j/=mp[j])if(v[i].empty()||v[i].back()!=mp[j])v[i].push_back(mp[j]);
}
bool chepre(int x){
	for(auto i:v[x]){
		int j=x/i;
		if(Z1[j]+j>=x)return false;
	}
	return true;
}
bool chesuf(int x){
	x=n-x;
	for(auto i:v[x]){
		int j=x/i;
		if(Z2[j]+j>=x)return false;
	}
	return true;
}
bool checkallsame(){
	for(int i=1;i<n;i++)if(s[i]!=s[0])return true;
	return false;
}
int main(){
	scanf("%s",s),n=strlen(s);
	ural(),fact();
	Zalgo(Z1),reverse(s,s+n),Zalgo(Z2);
	if(!checkallsame()){printf("%d\n1\n",n);return 0;}
	if(chepre(n)){printf("1\n1\n");return 0;}
	puts("2");
	for(int i=1;i<n;i++)res+=(chepre(i)&&chesuf(i));
	printf("%d\n",res);
	return 0;
} 