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 int long long

using namespace std;

const int N = 5e5 + 5;
const int base = 317;
const int mod = 1e9 + 2277;

string s;
int Hash[N], pw[N], n;
bool pref[N], suff[N];

void buildHash(){
    s = "#" + s;
    pw[0] = 1;
    for (int i = 1; i < N; i++) pw[i] = (pw[i - 1] * base) % mod;
    Hash[0] = 0;
    for (int i = 1; i <= n; i++) Hash[i] = (Hash[i - 1] * base + s[i]) % mod;
}

int get (int l, int r) {
    return ((1LL * Hash[r] - 1LL * Hash[l - 1] * pw[r - l + 1]) % mod + mod) % mod;
}

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin >> s;
    n = s.length();

    bool same = true;
    for (int i = 0; i < n; i++) if (s[i] != s[0]) same = false;
    if (same) {
        cout << n << "\n" << 1;
        return 0;
    }

    buildHash();
    memset(pref, 0, sizeof(pref));
    memset(suff, 0, sizeof(suff));

    for (int i = 1; i <= n; i++) {
        for (int j = i * 2; j <= n; j = j + i) {
            if (get(1, i) == get(j - i + 1, j)) pref[j] = true;
            else break;
        }
    }

    for (int i = 1; i <= n; i++) {
        for (int j = n - i * 2 + 1; j >= 1; j = j - i) {
            if (get(j, j + i - 1) == get(n - i + 1, n)) suff[j] = true;
            else break;
        }
    }

    if (!pref[n]) return cout << 1 << "\n" << 1, 0;
    int ans = 0;
    for (int i = 1; i < n; i++) if (!pref[i] && !suff[i + 1]) ans++;
    cout << 2 << "\n" << ans;
    return 0;
}
