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`).

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.

Examples

Input

aab


Output

1
1


Input

bcbc


Output

2
3


Input

ddd


Output

3
1
#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e5 + 5;
int n;
char s[maxn];
void kmp(char *s, vector<int> &nxt) {
	for(int i = 2, j = 0; i <= n; ++i) {
		while(j && s[j + 1] != s[i]) {
			j = nxt[j];
		}
		if(s[j + 1] == s[i]) {
			++j;
		}
		nxt[i] = j;
	}
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	cin >> s + 1;
	n = strlen(s + 1);
	vector<int> f;
	vector<int> g;
	f.resize(n + 1);
	g.resize(n + 1);
	kmp(s, f);
	for(int i = 1; i <= n; ++i) {
		f[i] = i - f[i];
	}
	reverse(s + 1, s + n + 1);
	kmp(s, g);
	for(int i = 1; i <= n; ++i) {
		g[i] = i - g[i];
	}
	if(f[n] == n || n % f[n] != 0) {
		cout << 1 << '\n' << 1 << '\n';
		return 0;
	}
	if(f[n] == 1) {
		cout << n << '\n' << 1 << '\n';
		return 0;
	}
	int ans = 0;
	for(int i = 1; i < n; ++i) {
		if((f[i] == i || i % f[i] != 0) && (g[n - i] == n - i || (n - i) % g[n - i] != 0)) {
			++ans;
		}
	}
	cout << 2 << '\n' << ans << '\n';
	return 0;
}