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<algorithm>
#include<string>
#include<iostream>
using namespace std;

typedef long long ll;
const ll MOD=1e9+7;
const ll base=1123;
const int MAXN=5e5+5;

string s;
int n;
ll pow[MAXN],Hash[MAXN];
bool l[MAXN],r[MAXN];

void Init(){
    s=' '+s;
    pow[0]=1;
    for(int i=1;i<=n;i++) pow[i]=(1ll*pow[i-1]*base)%MOD;
    for(int i=1;i<=n;i++) Hash[i]=(1ll*Hash[i-1]*base+s[i]-'a')%MOD;
}

ll get(int l,int r){
    return ((1ll*Hash[r]-1ll*Hash[l-1]*pow[r-l+1])%MOD+MOD)%MOD;
}

int main(){
    cin >> s;
    n=s.length();
    bool same=1;
    for(int i=1;i<s.length();i++) if(s[i]!=s[i-1]){same=0;break;}
    if(same) return printf("%d\n1",s.length()),0;
    Init();
    for(int i=1;i<=n;i++)
        for(int j=i*2;j<=n;j+=i){
            if(get(1,i)==get(j-i+1,j)) l[j]=1;
            else break;
        }
    for(int i=1;i<=n;i++)
        for(int j=n-i*2+1;j>=1;j-=i){
            if(get(j,j+i-1)==get(n-i+1,n)) r[j]=1;
            else break;
        }
    if(!l[n]) return printf("1\n1"),0;
    int ans=0;
    for(int i=1;i<n;i++) if(!l[i]&&!r[i+1]) ans++;
    printf("2\n%d",ans);
}
