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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

const int N = 550000;

vector<int> d[N];

char a[N], b[N];

int f[N], g[N];

void KMP(char *P, int n) {
	f[1] = 0;
	for (int i = 1, j = 0; i < n; i++) {
		while (j && P[i] != P[j]) j = f[j];
		if (P[i] == P[j]) j++;
		f[i+1] = j;
	}
}

void kmp(char *P, int n) {
	g[1] = 0;
	for (int i = 1, j = 0; i < n; i++) {
		while (j && P[i] != P[j]) j = g[j];
		if (P[i] == P[j]) j++;
		g[i+1] = j;
	}
}

bool ok(int st, int en) {
	int len = en - st + 1;
	if (st == 0) {
		if (f[en + 1] == 0) return 1;
		if (len % (len - f[en + 1]) == 0) return 0;
		return 1;
	} else {
		if (g[len - st] == 0) return 1;
		if (len % (len - g[len - st]) == 0) return 0;
		return 1;
	}
	return 1;
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	for (int i = 2; i < N; i++) for (int j = i; j < N; j += i) d[j].push_back(i);
	cin >> a;
	int len = strlen(a);
	for (int i = 0; i < len; i++) b[i] = a[i];
	reverse(b, b + len);
	KMP(a, len); kmp(b, len);
	if (ok(0, len - 1)) {
		cout << "1\n1\n";
		return 0;
	}
	bool flg = 1;
	for (int i = 1; i < len; i++) if (a[i] != a[0]) flg = 0;
	if (flg) {
		cout << len << "\n1\n";
		return 0;
	}
	int ans = 0;
//	cerr << ok(0, 0) << endl; //" " << ok(1, 3) << endl;
	for (int i = 0; i < len - 1; i++) if (ok(0, i) && ok(i + 1, len - 1)) {
		ans++;
	}
	cout << 2 << endl << ans << endl;
	return 0;
}