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<cstring>
#include<cmath>
#include<algorithm>
#define SF scanf
#define PF printf
#define MAXN 500010
using namespace std;
typedef long long ll;
ll pre[MAXN],las[MAXN],fsp[MAXN];
const ll MOD=1000000009ll;
int prel[MAXN],lasl[MAXN];
char s[MAXN];
int n,minlen;
void check(int x,int flag){
	if(flag==0){
		prel[x]=x;
		for(int i=1;i*i<=x;i++)
			if(x%i==0){
				if(((pre[x]-pre[i]*fsp[x-i])%MOD+MOD)%MOD==pre[x-i]%MOD){
					prel[x]=i;
					break;
				}
				if(((pre[x]-pre[x/i]*fsp[x-x/i])%MOD+MOD)%MOD==pre[x-x/i]%MOD){
					prel[x]=x/i;	
				}
			}
	}
	else{
		lasl[x]=x;
		for(int i=1;i*i<=x;i++)
			if(x%i==0){
				if(((las[x]-las[i]*fsp[x-i])%MOD+MOD)%MOD==las[x-i]%MOD){
					lasl[x]=i;	
					break;
				}
				if(((las[x]-las[x/i]*fsp[x-x/i])%MOD+MOD)%MOD==las[x-x/i]%MOD){
					lasl[x]=x/i;	
				}
			}
	}
}
int main(){
	SF("%s",s+1);
	n=strlen(s+1);
	fsp[0]=1;
	for(int i=1;i<=n;i++)
		fsp[i]=fsp[i-1]*12309097ll%MOD;
	for(int i=1;i<=n;i++)
		pre[i]=(pre[i-1]*12309097ll%MOD+(ll)(s[i]-'a'+1))%MOD;
	reverse(s+1,s+n+1);
	for(int i=1;i<=n;i++)
		las[i]=(las[i-1]*12309097ll%MOD+(ll)(s[i]-'a'+1))%MOD;
	for(int i=1;i<=n;i++){
		check(i,0);
		check(i,1);
	}
	if(prel[n]==1){
		PF("%d\n1",n);
		return 0;	
	}
	if(prel[n]==n){
		PF("1\n1");
		return 0;
	}
	int ans=0;
	
	for(int i=1;i<n;i++){
		if(prel[i]==i&&lasl[n-i]==n-i)
			ans++;
	}
	PF("2\n%d",ans);
}