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 ll long long
#define uint unsignextd long long
#define db double
#define ls x << 1
#define rs x << 1 | 1
#define pb push_back
#define mp make_pair
#define pii pair<int, int>
#define X first
#define Y second
#define pcc pair<char, char>
#define vi vector<int>
#define vl vector<ll>
#define rep(i, x, y) for(int i = x; i <= y; i ++)
#define rrep(i, x, y) for(int i = x; i >= y; i --)
#define eps 1e-9
using namespace std;
inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}
const int N = 500010 , mod = 1e9 + 7;
int n, nxt[N], next2[N];
char s[N];
bool is(int x, int *nxt)
{
    return nxt[x] && x % (x - nxt[x]) == 0;
}
 
int main()
{
    scanf("%s",s + 1);
    n = strlen(s + 1);
    int j = 0;
    rep(i, 2, n)
    {
        while(j && s[j + 1] != s[i]) j = nxt[j];
        nxt[i] = s[j + 1] == s[i] ? ++ j : j;
    }
    if(!is(n, nxt)) printf("1\n1\n");
    else if(nxt[n] == n - 1) printf("%d\n1\n",n);
    else
    {
        reverse(s + 1, s + 1 + n);
        j = 0;
        rep(i, 2, n)
	    {
	        while(j && s[j + 1] != s[i]) j = next2[j];
	        next2[i] = (s[j + 1] == s[i] ? ++ j : j);
	    }
        int ans = 0;
        rep(i, 1, n - 1) 
        	if(!is(i, nxt) && !is(n - i, next2)) ans ++;
        printf("2\n%d\n", ans);
    }
    return 0;
}
