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;

typedef long long ll;
const int N = 500005;
const int M = 1000000007;

int n;
char s[N];

int nxt[N], nxtr[N];
void getnxt(char *x, int m, int nxt[]) {
  int i = 0, j = -1; nxt[0] = -1;
  while (i < m) {
    while (j != -1 && x[i] != x[j]) { j = nxt[j]; }
    nxt[++i] = ++j;
  }
}

inline bool isloop(int len, int nxt[]) {
  return nxt[len] && len % (len - nxt[len]) == 0;
}

int main() {
  while (~scanf("%s", s)) {
    n = strlen(s);
    getnxt(s, n, nxt);
    if (!isloop(n, nxt)) { printf("1\n1\n"); continue; }
    if (nxt[n] == n - 1) { printf("%d\n1\n", n); continue; }
    reverse(s, s + n);
    getnxt(s, n, nxtr);
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      ans += (!isloop(i, nxt) && !isloop(n - i, nxtr));
    }
    printf("2\n%d\n", ans);
  }
}
