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 A[2][500005];
string S[2];
void Z(int x){
	A[x][0] = S[x].size();
	int i=1,j=0;
	while(i<S[x].size()){
		while(i+j<S[x].size() && S[x][j]==S[x][i+j])j++;
		A[x][i]=j;
		if(j==0){
			i++; continue;
		}
		int k = 1;
		while(i+k<S[x].size() && k+A[x][k]<j) A[x][i+k]=A[x][k],++k;
		i+=k;j-=k;
	}
}
vector<int>v[500005];
bool bad[500005],bad2[500005];
int main(){
	cin >> S[0];
	reverse(S[0].begin(),S[0].end());
	S[1] = S[0];
	reverse(S[0].begin(),S[0].end());
	Z(0); Z(1);
	for(int i=1;i<500005;i++)for(int j=2*i;j<500005;j+=i)v[j].push_back(i);
	for(int i=0;i<S[0].size();i++){
		for(int x=0;x<v[i+1].size();x++){
			int a = v[i+1][x];
			if(A[0][a] >= i+1-a){
				bad[i] = 1; break;
			}
		}
	}
	for(int i=0;i<S[1].size();i++){
		for(int x=0;x<v[i+1].size();x++){
			int a = v[i+1][x];
			if(A[1][a] >= i+1-a){
				bad2[i] = 1; break;
			}
		}
	}
	if(!bad[S[0].size()-1]){
		cout << 1 << endl << 1 << endl;
	}
	else{
	    int ans=0;
		for(int i=1;i<S[0].size();i++){
			if(!bad[i-1] && !bad2[S[0].size()-i-1]) ans++;
		}
		if(!ans) cout << S[0].size() << endl << 1 << endl;
		else cout << 2 << endl << ans << endl;
	}
}