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 N = 500005;

int n, p[2][N];
char s[N];

int main() {
    scanf("%s", s);
    n = strlen(s);
    for (int id = 0; id < 2; ++id) {
        int j = 0;
        for (int i = 1; i < n; ++i) {
            while (j > 0 && s[i] != s[j]) j = p[id][j - 1];
            if (s[i] == s[j]) ++j;
            p[id][i] = j;
        }
        reverse(s, s + n);
    }
    int len = n - p[0][n - 1];
    if (len == n || n % len != 0) {
        cout << 1 << endl << 1 << endl;
        return 0;
    }
    if (len == 1) {
        cout << n << endl << 1 << endl;
        return 0;
    }
    int cnt = 0;
    for (int i = 0; i < n - 1; ++i) {
        int pos = n - (i + 1) - 1;
        int len[2] = {i + 1 - p[0][i], n - i - 1 - p[1][pos]};
        if ((len[0] == i + 1 || (i + 1) % len[0] != 0) && ((n - i - 1) % len[1] != 0 || len[1] == n - i - 1)) {
            ++cnt;
        }
    }
    cout << 2 << endl << cnt << endl;
    return 0;
}
