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 < n; i++)
using namespace std;
int qmin(int a, int b){
	return (a < b) ? a : b;
}

int n;
char s[500500]; 
int Z[2][500500];
bool nok[500500];

void Zalgo(int dir){
	int l = 0, r = 0;
	Z[dir][0] = n;
	for(int i = 1; i < n; i++){
		if(i <= r) Z[dir][i] = qmin(Z[dir][i-l], r-i+1);
		while(i + Z[dir][i] < n && s[i + Z[dir][i]] == s[Z[dir][i]]) Z[dir][i]++;
		if(i+Z[dir][i]-1 > r){
			l = i;
			r = i+Z[dir][i]-1;
		}
	}
}

int main(){
	scanf("%s", s);
	n = strlen(s);
	bool all_same = 1;
	rep(i, n){
		if(s[i] != s[0]) all_same = 0;
	}
	if(all_same){
		cout << n << endl << "1" << endl;
		return 0;
	}
	Zalgo(0);
	for(int i = 0; i + i < n; i++){
		swap(s[i], s[n-i-1]); 
	}
	Zalgo(1);
	for(int i = 1; i <= n; i++){
		for(int j = 2; i*j <= n; j++){
			if(Z[0][i*(j-1)] >= i){
				nok[i*j] = 1;
			} else break;
		} 
		for(int j = 2; i*j <= n; j++){
			if(Z[1][i*(j-1)] >= i){
				nok[n - i*j] = 1;
			} else break;
		}
	}
	if(!nok[n]){
		cout << "1" << endl << "1" << endl; 
	} else {
		int cnt = 0;
		for(int i = 1; i < n; i++){
			if(!nok[i]) cnt++;
		}
		cout << "2" << endl << cnt << endl;
	}
	return 0;
}