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 <vector>
#include <iostream>
#include <unordered_map>
#include <map>
#include <iomanip>
#include <functional>
#include <algorithm>
#include <cassert>
#include <cmath>
#include <string>
#include <sstream>
using namespace std;

#ifndef MDEBUG
#define NDEBUG
#endif

#define x first
#define y second
#define ll long long
#define d double
#define ld long double
#define pii pair<int,int>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pll pair<ll,ll>
#define pss pair<string,string>
#define psi pair<string,int>
#define pis pair<int,string>
#define psl pair<string,ll>
#define pls pair<ll,string>
#define wh(x) (x).begin(),(x).end()
#define ri(x) int x;cin>>x;
#define rii(x,y) int x,y;cin>>x>>y;
#define rl(x) ll x;cin>>x;
#define rv(v) for(auto&_cinv:v) cin>>_cinv;
#define wv(v) for(auto&_coutv:v) cout << _coutv << ' '; cout << endl;
#define ev(v) for(auto&_cerrv:v) cerr << _cerrv << ' '; cerr << endl;
#define MOD 1000000007

namespace std { 
template<typename T,typename U>struct hash<pair<T,U>>{hash<T>t;hash<U>u;size_t operator()(const pair<T,U>&p)const{return t(p.x)^(u(p.y)<<7);}};
}
// auto fraclt = [](auto&a,auto&b) { return (ll)a.x * b.y < (ll)b.x * a.y; };
template<typename T,typename F>T bsh(T l,T h,const F&f){T r=-1,m;while(l<=h){m=(l+h)/2;if(f(m)){l=m+1;r=m;}else{h=m-1;}}return r;}
template<typename T,typename F>T bsl(T l,T h,const F&f){T r=-1,m;while(l<=h){m=(l+h)/2;if(f(m)){h=m-1;r=m;}else{l=m+1;}}return r;}
template<typename T> T gcd(T a,T b) { if (a<b) swap(a,b); return b?gcd(b,a%b):a; }
template<typename T> void fracn(pair<T,T>&a) {auto g=gcd(a.x,a.y);a.x/=g;a.y/=g;}
template<typename T> struct Index { int operator[](const T&t){auto i=m.find(t);return i!=m.end()?i->y:m[t]=m.size();}int s(){return m.size();} unordered_map<T,int> m; };


pii isgood(string W) {
    int N = W.size();
    if (N == 1) return {1,1};

    bool diff = false;
    for (int i = 1; i < N && !diff; i++) {
        diff |= (W[0] != W[i]); 
    }
    if (!diff) return {N,1};
    
    for (int i = 2; i < N; i++) {
        if (N % i != 0) continue;
        bool diff = false;
        for (int j = i; j < N && !diff; j+=i) {
            for (int k = 0; k < i && !diff; k++) {
                diff |= W[k] != W[j+k];
            }
        }
        if (!diff) {
            int g = 0;
            for (int j = 1; j < N; j++) {
                pii a = isgood(W.substr(0,j));
                pii b = isgood(W.substr(j,N));
                if (a.x == 1 && b.x == 1) ++g;
            }


            return {2,g};
        }
    }
    return {1,1};
}

int main(int,char**) {
    ios_base::sync_with_stdio(false);

    string W; cin >> W;
	if (W.size() > 4000) return 0;
    pii x = isgood(W);
    cout << x.x << endl << x.y << endl;
}