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>

#define ln                '\n'
#define all(dat)           dat.begin(), dat.end()
#define loop(i, to)        for (int i = 0; i < to; ++i)
#define cont(i, to)        for (int i = 1; i <= to; ++i)
#define circ(i, fm, to)    for (int i = fm; i <= to; ++i)
#define foreach(i, dat)    for (__typeof(dat.begin()) i = dat.begin(); i != dat.end(); ++i)

typedef long long          num;

using namespace std;

int n, d, t, ans;
string s, rs;
vector<int> l, rl;

void inline kmp_init(string s, vector<int> &l) {
	l.resize(s.size()), fill(all(l), 0);
	for (int i = 1, k = 0; i < s.size(); ++i) {
		for (; k && s[i] != s[k]; k = l[k - 1]);
		l[i] = k += s[i] == s[k];
	}
}

int main() {
	ios::sync_with_stdio(0);
	cin >> s;
	rs = s, reverse(all(rs));
	n = (int) s.size();
	kmp_init(s, l);
	kmp_init(rs, rl);
	reverse(all(rl));
	d = n - l[n - 1];
	if (d == n || n % d) {
		cout << 1 << ln << 1 << ln;
	} else if (d > 1) {
		loop (i, n - 1) {
			int l1 = i + 1, l2 = n - l1, dd1 = l1 - l[i], dd2 = l2 - rl[i + 1];
			ans += (l1 == dd1 || l1 % dd1) && (l2 == dd2 || l2 % dd2);
		}
		cout << 2 << ln << ans << ln;
	} else {
		cout << n << ln << 1 << ln;
	}
}