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>
typedef long long int ll;
typedef long double ld;
#define pb push_back
#define pii pair < int, int >
#define F first
#define S second
#define int  long long int
#define sync ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#pragma GCC optimize ("Ofast")
#pragma GCC optimize ("unroll-loops")
using namespace std;
/// khodaya komak kon

/// ba z_func hatman dafe baad bezan
const int MAX_N = 5e5 + 5;
char a[MAX_N];
int N, pre[MAX_N], nxt[MAX_N];
bool check(int pos) {
	bool f1 = (!pre[pos]) || (pos % (pos - pre[pos]));
	bool f2 = (nxt[pos + 1] == N + 1) || (N - pos) % (N - pos - (N - nxt[pos + 1] + 1));
	if (f1 && f2) return 1;
	else return 0;
}
int32_t main () {
	scanf("%s", a + 1);
	N = strlen(a + 1);
	pre[1] = 0;
	for (int i = 2, j = 0; i <= N; i++) {
		while (j && a[j + 1] != a[i]) j = pre[j];
		if (a[j + 1] == a[i]) ++j;
		pre[i] = j;
	}
	nxt[N] = N + 1;
	for (int i = N - 1, j = N + 1; i; i--) {
		while (j <= N && a[j - 1] != a[i]) j = nxt[j];
		if (a[j - 1] == a[i]) --j;
		nxt[i] = j;
	}
	if (!pre[N] || N % (N - pre[N])) return puts("1\n1") & 0;
	if (pre[N] == N - 1) return printf("%d\n1\n", N) & 0;
	int ans = 0;
	for (int i = 1; i < N; i++) ans += check(i);
	printf("2\n%d\n", ans);
    return 0;
}
