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 <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using std::min; using std::max;
using std::swap; using std::sort;
typedef long long ll;

template<typename T>
void read(T &x) {
	int flag = 1; x = 0; char ch = getchar();
	while(ch < '0' || ch > '9') { if(ch == '-') flag = -flag; ch = getchar(); }
	while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
	x *= flag;
}

const int _ = 5e5 + 10, P = 1e9 + 7;
int n, p[_], s[_]; char c[_];

int chk0(int r) {
	int len = r, now = len % (len - p[r]);
	if(!now) return len / (len - p[r]) == 1;
	return 1;
}

int chk1(int l) {
	int len = n - l + 1, now = len % (len - s[l]);
	if(!now) return len / (len - s[l]) == 1;
	return 1;
}

int main () {
	scanf("%s", c + 1), n = strlen(c + 1);
	for(int i = 2; i <= n; ++i) {
		int j = p[i - 1];
		while(j && c[j + 1] != c[i]) j = p[j];
		j += c[j + 1] == c[i], p[i] = j;
	}
	std::reverse(c + 1, c + n + 1);
	for(int i = 2; i <= n; ++i) {
		int j = s[i - 1];
		while(j && c[j + 1] != c[i]) j = s[j];
		j += c[j + 1] == c[i], s[i] = j;
	}
	std::reverse(s + 1, s + n + 1);
	std::reverse(c + 1, c + n + 1);
	int now = n - p[n];
	if(now == 1) return printf("%d\n%d\n", n, 1), 0;
	if(n % now == 0 && n / now > 1) {
		printf("%d\n", 2);
		int ret = 0;
		for(int i = 1; i < n; ++i)
			if(chk0(i) && chk1(i + 1)) ++ret;
		printf("%d\n", ret);
	} else printf("%d\n%d\n", 1, 1);
	return 0;
} 
