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 <iostream>
#include <sstream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <cstring>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <numeric>
#include <utility>
#include <iomanip>
#include <algorithm>
#include <functional>
using namespace std;

#define COUT(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl
#define EACH(i, s) for (__typeof__((s).begin()) i = (s).begin(); i != (s).end(); ++i)

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<class T1, class T2> ostream& operator << (ostream &s, pair<T1,T2> P)
{ return s << '<' << P.first << ", " << P.second << '>'; }
template<class T> ostream& operator << (ostream &s, vector<T> P)
{ for (int i = 0; i < P.size(); ++i) { if (i > 0) { s << " "; } s << P[i]; } return s; }
template<class T> ostream& operator << (ostream &s, vector<vector<T> > P)
{ for (int i = 0; i < P.size(); ++i) { s << endl << P[i]; } return s << endl; }
template<class T1, class T2> ostream& operator << (ostream &s, map<T1,T2> P)
{ EACH(it, P) { s << "<" << it->first << "->" << it->second << "> "; } return s << endl; }

const int MAX = 550000;
vector<long long> divisor(long long n) {
    vector<long long> res;
    for (long long i = 1LL; i*i <= n; ++i) {
        if (n%i == 0LL) {
            res.push_back(i);
            long long temp = n/i;
            if (i != temp) res.push_back(temp);
        }
    }
    sort(res.begin(), res.end());
    return res;
}

string str;


bool mae[MAX], ato[MAX], submae[MAX], subato[MAX];
void check() {
    for (int i = 0; i < MAX; ++i) mae[i] = ato[i] = false;
    for (int d = 1; d <= (int)str.size()/2; ++d) {
        for (int j = 0; j * d < str.size(); ++j) {
            bool ok = true;
            for (int k = 0; k < d; ++k) {
                if (str[d*j + k] != str[d*(j+1) + k]) {
                    ok = false;
                }
            }
            if (ok) {
                mae[d*(j+1)] = true;
                //cout << d * (j+2) << endl;
            }
        }
    }
    reverse(str.begin(), str.end());
    for (int d = 1; d <= (int)str.size()/2; ++d) {
        for (int j = 0; j * d < str.size(); ++j) {
            bool ok = true;
            for (int k = 0; k < d; ++k) {
                if (str[d*j + k] != str[d*(j+1) + k]) {
                    ok = false;
                }
            }
            if (ok) {
                ato[d*(j+1)] = true;
            }
        }
    }
}

int main() {
    while (cin >> str) {
        int n = str.size();
        vector<long long> divs = divisor(n);
        int syu = n;
        for (int i = 0; i < divs.size(); ++i) {
            int d = divs[i];
            bool ok = true;
            for (int j = 0; j + d < n; ++j) {
                if (str[j] != str[j+d]) {
                    ok = false;
                }
            }
            if (ok) {
                chmin(syu, d);
            }
        }
        if (syu == n) cout << 1 << endl << 1 << endl;
        else if (syu == 1) cout << n << endl << 1 << endl;
        else {
            check();
            
            int num = 0;
            for (int i = 1; i < syu; ++i) {
                if (mae[i]) ++num;
                if (ato[i]) ++num;
            }
            //COUT(num);
            if (n/syu > 2) cout << 2 << endl << n-1 - (n/syu - 1) - num << endl;
            else cout << 2 << endl << n-1-num << endl;
        }
    }
}