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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MN = 500500;
ll P = 53;
ll MA = 1e9+7,MB = 1e9+9;
ll sa[MN],sb[MN];
ll ra[MN],rb[MN];
char st[MN];
ll n;
ll hs(ll s, ll l) {
	ll ha = ra[s+l]-sa[l]*ra[s];
	ll hb = rb[s+l]-sb[l]*rb[s];
	ha = ((ha%MA)+MA)%MA;
	hb = ((hb%MB)+MB)%MB;
	return (ha*MB)+hb;
}
bitset<MN> pr,su;
int main() {
	scanf("%s",st);
	n = strlen(st);
	bool ol = true;
	for(int i=0;i<n;i++) {
		if(st[i] != st[0]) {ol = false;break;}
	}
	if(ol) {
		printf("%d\n1\n",n);
		return 0;
	}
	sa[0] = sb[0] = 1;
	ra[0] = rb[0] = 0;
	for(int i=0;i<n;i++) {
		sa[i+1] = (sa[i]*P)%MA;
		sb[i+1] = (sb[i]*P)%MB;
		ra[i+1] = (P*ra[i]+st[i]-'a'+1)%MA;
		rb[i+1] = (P*rb[i]+st[i]-'a'+1)%MB;
	}
	pr.set();su.set();
	for(int i=1;i<=n;i++) {
		ll val = hs(0,i);
		for(int j=1;i*(j+1)<=n;j++) {
			if(val != hs(i*j,i)) {
				break;
			}
			pr.reset(i*(j+1));
		}
		val = hs(n-i,i);
		for(int j=1;i*(j+1)<=n;j++) {
			if(val != hs(n-i*(j+1),i)) {
				break;
			}
			su.reset(n-i*(j+1));
		}
	}
	if(pr[n]) {
		printf("1\n1\n");
		return 0;
	}
	/*
	for(int i=1;i<n;i++) {
		printf("%d",pr[i]?1:0);
	}
	printf("\n");
	for(int i=1;i<n;i++) {
		printf("%d",su[i]?1:0);
	}
	printf("\n");
	*/
	ll tot = 0;
	for(int i=1;i<n;i++) {
		tot += pr[i]*su[i];
	}
	printf("2\n");
	printf("%lld\n",tot);
}