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 mp make_pair
#define pb push_back
#define IOS ios::sync_with_stdio(false); cin.tie(0);
#define sd(n) scanf("%d",&n)
#define sl(n) scanf("%lld",&n)
#define pd(n) printf("%d",n)
#define pl(n) printf("%lld",n)
#define endc printf("\n")

using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;

const int maxn = 5e5 + 7; 
//const int md = 1e9 + 7;

int f[maxn][2];
int kmp[maxn];

void calc(string s, int n, int p) {
	fill(kmp+1,kmp+n+1,0); int q = 0; kmp[1] = q; f[1][p] = 1;
	for (int i = 2; i <= n; i++) {
		while (q > 0 && s[q+1] != s[i]) q = kmp[q];
		if (s[q+1] == s[i]) q++; kmp[i] = q;
		int d = i - kmp[i];
		if (i % d == 0) f[i][p] = i/d;
		else f[i][p] = 1;
	}
}

int main() {
	string s; cin>>s; string t = s; reverse(t.begin(), t.end());
	int n = s.size(); s = " " + s; t = " " + t; calc(s,n,0); calc(t,n,1);
	if (f[n][0] == n) {
		pd(n); endc; pd(1); endc; return 0;
	} else if (f[n][0] == 1) {
		pd(1); endc; pd(1); endc; return 0;
	}
	int ans = 0;
	for (int i = 1; i < n; i++) {
		if (f[i][0] == 1 && f[n-i][1] == 1) ans++;
	}
	pd(2); endc; pd(ans); endc; return 0;
}
