Problem: Kenus, the organizer of International Euclidean Olympiad, is seeking a pair of two integers that requires many steps to find its greatest common divisor using the Euclidean algorithm.

You are given Q queries. The i-th query is represented as a pair of two integers X_i and Y_i, and asks you the following: among all pairs of two integers (x,y) such that 1 ≤ x ≤ X_i and 1 ≤ y ≤ Y_i, find the maximum Euclidean step count (defined below), and how many pairs have the maximum step count, modulo 10^9+7.

Process all the queries. Here, the Euclidean step count of a pair of two non-negative integers (a,b) is defined as follows:

* (a,b) and (b,a) have the same Euclidean step count.
* (0,a) has a Euclidean step count of 0.
* If a > 0 and a ≤ b, let p and q be a unique pair of integers such that b=pa+q and 0 ≤ q < a. Then, the Euclidean step count of (a,b) is the Euclidean step count of (q,a) plus 1.

Constraints

* 1 ≤ Q ≤ 3 × 10^5
* 1 ≤ X_i,Y_i ≤ 10^{18}(1 ≤ i ≤ Q)

Input

The input is given from Standard Input in the following format:


Q
X_1 Y_1
:
X_Q Y_Q


Output

For each query, print the maximum Euclidean step count, and the number of the pairs that have the maximum step count, modulo 10^9+7, with a space in between.

Examples

Input

3
4 4
6 10
12 11


Output

2 4
4 1
4 7


Input

10
1 1
2 2
5 1000000000000000000
7 3
1 334334334334334334
23847657 23458792534
111111111 111111111
7 7
4 19
9 10


Output

1 1
1 4
4 600000013
3 1
1 993994017
35 37447
38 2
3 6
3 9
4 2

Solution (in CPP):

Solution:
// Crt. 2020-01-10  08:56:59
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long ll;
const int Mod = 1e9 + 7;
const int lim = 90;

ll n, m, fib[200];

inline ll calc(ll a, ll b) {
    ll res; res = (m - a) / b + 1, res %= Mod;
    if (a <= n) res += (n - a) / b + 1, res %= Mod;
    return res;
}

void work(int K) {
    ll res = 0; res += calc(fib[K + 1], fib[K]), res %= Mod;
    for (int i = 2; i < K; ++i) {
        ll x = fib[i + 2], y = fib[i];
        ll a = x * fib[K - i + 1] + y * fib[K - i];
        ll b = x * fib[K - i] + y * fib[K - i - 1];
        if (a > m || b > n || !a || !b) continue;
        res += calc(a, b), res %= Mod;
    }
    printf("%d %lld\n", K - 1, res);
}

int main(){
    fib[0] = 0, fib[1] = 1;
    for (int i = 2; i <= lim; ++i)
        fib[i] = fib[i - 1] + fib[i - 2];

    int Q; cin >> Q;
    while (Q--) {
        scanf("%lld%lld", &n, &m);
        if (n > m) swap(n, m); int K = 0;
        if (n == 1) { printf("1 %lld\n", (m) % Mod); continue;}
        if (n + m == 4) { puts("1 4"); continue;}
        for (int i = 1; i < lim; ++i)
            if (fib[i + 1] <= m && fib[i] <= n)
                K = i;
        work(K);
    }
    return 0;
}