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:
#include <bits/stdc++.h>
using namespace std;
#define int long long
int read() {
    int x; scanf("%lld", &x); return x;
}
#define xx first
#define yy second
#define pb push_back
typedef pair<int, int> pir;
const int mod = 1000000007;

vector<pir> f, s[100];
int fib[100];
int find(int a, int b) {
	int ka = upper_bound(fib + 1, fib + 91, a) - fib;
	int kb = upper_bound(fib + 1, fib + 91, b) - fib;
	if (ka > kb + 1) ka = kb + 1;
	if (ka == kb) kb--;
	return kb - 1;
}

signed main() {
    f.pb(pir(1, 1));
    f.pb(pir(3, 2));
    f.pb(pir(4, 3));
    s[2].pb(pir(3, 2));
    for (int i = 3; i <= 90; i++) {
        f.pb(pir(f[i - 1].xx + f[i - 2].xx, f[i - 1].yy + f[i - 2].yy));
        for (int j = 0; j < i - 2; j++) {
            s[i].push_back(pir(s[i - 1][j].xx + s[i - 1][j].yy, s[i - 1][j].xx));
        }
        s[i].pb(pir(f[i - 1].xx + f[i - 1].yy, f[i - 1].xx));
    }
    fib[0] = fib[1] = 1;
    for (int i = 2; i <= 90; i++)
    	fib[i] = fib[i - 1] + fib[i - 2];
    	
    int Q = read();
    while (Q--) {
    	int A = read(), B = read(), ans = 0;
		if (A < B) swap(A, B);
		if (B == 1) { printf("1 %lld\n", A % mod); continue; }
		if (A == 2 && B == 2) { puts("1 4"); continue; }
    	int k = find(A, B);
    	printf("%lld ", k);
    	for (int i = 0, _ = s[k].size(); i < _; i++) {
    		int x = s[k][i].xx, y = s[k][i].yy;
    		if (y <= B && x <= A) {
    			ans = (ans + (A - x) / y + 1) % mod;
    		}
    		if (y <= A && x <= B) {
    			ans = (ans + (B - x) / y + 1) % mod;
    		}
    	}
    	if (f[k].xx <= A && f[k].yy <= B)
    		ans = (ans + 1) % mod;
    	if (f[k].xx <= B && f[k].yy <= A)
    		ans = (ans + 1) % mod;
    	printf("%lld\n", ans);
    }
}