Problem:
Score : 900 points

Problem Statement

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).

Partial Score

400 points will be awarded for passing the test set satisfying 1 \leq |w| \leq 4000.

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.

Sample Input 1

aab

Sample Output 1

1
1

Sample Input 2

bcbc

Sample Output 2

2
3

In this case, there are 3 best representations of 2 elements:

(b,cbc)

(bc,bc)

(bcb,c)

Sample Input 3

ddd

Sample Output 3

3
1

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#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];
bool ispr[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);
	rep(i, n+1) ispr[i] = 1; 
	for(int i = 1; i <= n; i++){
		if(i > 1) for(int j = i+i; j <= i*i && j <= n; j += i) ispr[j] = 0;
		if(ispr[i] && (n % i) == 0){
			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;
}