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 + 7;
int nex[MAXN], nex2[MAXN];
void GetNext(char *p, int *nex)
{
    int m = strlen(p);
    memset(nex, 0, sizeof(nex));
    nex[0] = -1;
    int k = -1, j = 0;
    while (j < m)
    {
        if (k == -1 || p[j] == p[k]) nex[++j] = ++k;
        else k = nex[k];
    }
}
char s[MAXN], t[MAXN];
int main()
{
    scanf("%s", s); int len = strlen(s);
    for (int i = 0; i < len; i++) t[i] = s[len - 1 - i]; t[len] = '\0';
    GetNext(s, nex);
    int pd = 0;
    if (nex[len] && (len % (len - nex[len]) == 0)) pd = len - nex[len];
    if (pd == 0) printf("1\n1\n");
    else if (pd == 1) printf("%d\n1\n", len);
    else 
    {
        int ans = 0;
        GetNext(t, nex2);
        for (int i = 1; i < len; i++)
        {
            if (nex[i] && (i % (i - nex[i]) == 0)) continue;
            int j = len - i;
            if (nex2[j] && (j % (j - nex2[j]) == 0)) continue;
            ans++;
        }
        printf("2\n%d\n", ans);
    }
    return 0;
}