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 <algorithm>
#include <vector>
#include <map>
#include <string>
#include <cmath>
#include <iomanip>
#include <queue>
#include <cassert>

using namespace std;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#define SORT(c) sort((c).begin(), (c).end())

typedef long long ll;
typedef pair<int, int> P;
typedef vector<int> V;
typedef map<int, int> M;

const ll INF = 1e18;
const ll MOD = 1e9 + 7;

vector<int> prime(int n)
{
    vector<bool> num;
    vector<int> p;

    for (int i = 0; i <= n + 10; i++)
        num.push_back(true);

    FOR(i, 2, n)
    {
        if (!num[i])
            continue;

        p.push_back(i);

        for (int j = i * 2; j <= n; j += i)
        {
            num[j] = false;
        }
    }

    return p;
}

int main()
{
    string w;
    cin >> w;

    bool same = true;
    REP(i, w.size() - 1)
    if (w[i] != w[i + 1])
        same = false;

    if (same)
    {
        cout << w.size() << endl
             << 1 << endl;
        return 0;
    }

    int div = 0;

    for (int i = w.size() - 1; i > 1; i--)
    {
        if (w.size() % i)
            continue;

        string p = w.substr(0, w.size() / i);

        string c = "";
        REP(j, i)
        c += p;

        if (c == w)
        {
            div = i;
            break;
        }
    }

    if (div == 0)
    {
        cout << 1 << endl
             << 1 << endl;
        return 0;
    }

    int size = w.size() / div;

    int res = (size - 1) * (div);
    res += div == 2;

    V p = prime(size);

    //左側チェック
    int one = 0;
    REP(i, size - 1)
    {
        if (w[i] != w[i + 1])
            break;
        res--;
        one++;
    }

    for (int i : p)
    {
        if (one >= i)
            continue;

        string p = w.substr(0, i);
        int cnt = 2;
        string c = p + p;

        while (true)
        {
            if (p != w.substr(0, cnt * i))
                break;

            res--;
            cnt++;
            c += p;
        }
    }

    reverse(w.begin(), w.end());

    //右側チェック

    one = 0;
    REP(i, size - 1)
    {
        if (w[i] != w[i + 1])
            break;
        res--;
        one++;
    }

    for (int i : p)
    {
        if (one >= i)
            continue;

        string p = w.substr(0, i);
        int cnt = 2;
        string c = p + p;

        while (true)
        {
            if (p != w.substr(0, cnt * i))
                break;

            res--;
            cnt++;
            c += p;
        }
    }

    cout << 2 << endl
         << res << endl;

    return 0;
}