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 MOD 1000000007
#define int long long
#define INF 1061109567
#define pb push_back
#define in(s) freopen(s,"r",stdin);
#define out(s) freopen(s,"w",stdout);
#define fi first
#define se second
#define bw(i,r,l) for (int i=r-1;i>=l;i--)
#define fw(i,l,r) for (int i=l;i<r;i++)
#define fa(i,x) for (auto i:x)
using namespace std;
const int N=5e5+5;
string s;
int n,pw[N];
bool bad1[N],bad2[N];
bool check=true;
struct hash {
	int hs[N]; //This hash hashes lowercase - letters only strings
	void build(string s) {fw (i,0,s.length()) hs[i]=((i?hs[i-1]*26:0)+s[i]-'a')%MOD;}
	int get(int l,int r) {
		int ans=hs[r]-(l?(hs[l-1]*pw[r-l+1]%MOD):0);
		if (ans<0) ans+=MOD;
		return ans;
	}
} h;
signed main() {
	//in("aome.inp");
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	cin>>s; n=s.length(); s='~'+s;
	fw (i,1,n+1) if (s[i]!=s[1]) check=false;
	if (check) {
		cout<<n<<"\n"<<1;
		return 0;
	}
	pw[0]=1;
	fw (i,1,N) pw[i]=pw[i-1]*26%MOD;
	h.build(s);
	memset(bad1,false,sizeof(bad1));
	memset(bad2,false,sizeof(bad2));
	fw (i,1,n+1) {
		for (int j=i+i;j<=n;j+=i) if (h.get(j-i+1,j)==h.get(1,i)) bad1[j]=true; else break;
	}
	fw (i,1,n+1) {
		for (int j=n-i-i+1;j>0;j-=i) if (h.get(j,j+i-1)==h.get(n-i+1,n)) bad2[j]=true; else break;
	}
	if (!bad1[n]) {cout<<"1\n1"; return 0;}
	int cnt=0;
	fw (i,1,n) if (!bad1[i]&&!bad2[i+1]) cnt++;
	cout<<2<<"\n"<<cnt;
	return 0;
}