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 <algorithm>
#include <iostream>
#include <vector>
using namespace std;
const long long mod = 1e9+7;

vector<int> Z_algorithm(string S){
  int c = 0, n = S.length();
  vector<int> Z(n,0);
  for(int i = 1; i < n; ++i){
    int l = i - c;
    if(i + Z[l] < c + Z[c]){
      Z[i] = Z[l];
    }else{
      int j = max(0, c + Z[c] - i);
      while(i + j < n && S[j] == S[i+j]) ++j;
      Z[i] = j;
      c = i;
    }
  }
  Z[0] = n;
  return Z;
}

int main(){
  string w;
  cin >> w;
  vector<int> Z = Z_algorithm(w);
  reverse(w.begin(), w.end());
  vector<int> Z_rev = Z_algorithm(w);
  int l = w.length(), t = 0;
  vector<bool> V(l,true);
  for(int i = 1; i < l; ++i){
    int x = i;
    while(x <= Z[i]) x += i, V[x] = false;
    //if(Z[i] >= i) V[2*i] = false;
    if(Z[i] < l-i) continue;
    if(not t) t = i;
  }
  //for(int i = 0; i < l; ++i) cerr << V[i] << " ";cerr << endl;
  for(int i = 1; i < l; ++i){
    int x = i;
    while(x <= Z_rev[i]) x += i, V[l-x] = false;
    //if(Z_rev[i] >= i) V[l-2*i] = false;
  }
  //for(int i = 0; i < l; ++i) cerr << V[i] << " ";cerr << endl;
  //cerr << "t = " << t << endl;
  if(not t or l%t){
    cout << 1 << endl << 1 << endl;
    return 0;
  }
  if(t == 1){
    cout << l << endl << 1 << endl;
    return 0;
  }
  cout << 2 << endl;
  long long ans = 0;
  for(int i = 1; i < l; ++i){
    ans += V[i];
  }
  
  cout << ans << endl;
}
