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 rep(i, x, y) for (int i = x; i <= y; i++)
using namespace std;

const int N = 1e6 + 10;
int pre[N], n, sub[N];
char s[N], t[N];

bool chkpre(int x) {
    return x % (x - pre[x]) > 0 || !pre[x];
}

bool chksub(int x) {
    return x % (x - sub[x]) > 0 || !sub[x];
}

void getfail(char *s, int *nxt) {
    nxt[0] = nxt[1] = 0;
    int j = 0;
    rep(i, 1, n - 1) {
        while(j && s[i] != s[j]) j = nxt[j];
        nxt[i + 1] = (s[i] == s[j] ? ++j : 0);
    }
}

int main() {
    scanf("%s", s);
    n = strlen(s);
    rep(i, 0, n - 1) t[i] = s[n - i - 1];
    t[n] = '\0';

    getfail(s, pre), getfail(t, sub);
    
    int flag = chkpre(n) ? 1 : 2;
    bool f = 1;
    rep(i, 1, n - 1) if (s[i] != s[i - 1]) { f = 0; break; }
    if (f) flag = n;
    printf("%d\n", flag);
    if (flag != 2) { puts("1"); return 0; }

    int ans = 0;
    rep(i, 0, n - 1) {
        int x = n - i - 1;
        if (chkpre(i + 1) && chksub(x)) ans++;
    }
    printf("%d\n", ans);
    return 0;
}
