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;

typedef long long ll;
typedef pair<ll, ll> pll;

const int mod = 1000*1000*1000+7;

ll fibo[90];
vector<vector<pll> > prec;

ll X, Y;

void main2() {
    scanf("%lld %lld", &X, &Y);
    if(X > Y) swap(X, Y);

    if(X == 1 && Y == 1) {
        printf("1 1\n");
        return;
    }

    int K;
    for(K = 1; K < 90; K++) {
        if(fibo[K] <= X && fibo[K + 1] <= Y) continue;
        K--;
        break;
    }

    if(K == 1) {
        printf("1 %d\n", (2*X + Y - 2) % mod);
        return;
    }

    int cnt = 0;
    for(int i = 0; i < prec[K - 1].size(); i++) {
        pll t = prec[K - 1][i];

        if(t.second <= X && t.first + t.second <= Y) {
            cnt += (Y - t.first) / t.second % mod;
            cnt %= mod;
        }
        if(t.second <= Y && t.first + t.second <= X) {
            cnt += (X - t.first) / t.second % mod;
            cnt %= mod;
        }
    }

    printf("%d %d\n", K, cnt);
}

int TC;
int main() {
    fibo[0] = 1;
    fibo[1] = 1;
    for(int i = 2; i < 90; i++) {
        fibo[i] = fibo[i - 1] + fibo[i - 2];
    }

    prec.resize(90);
    prec[1].push_back(pll(1, 2));
    prec[1].push_back(pll(1, 3));

    for(int i = 1; i < 90; i++) {
        for(int j = 0; j < prec[i - 1].size(); j++) {
            pll t = prec[i - 1][j];

            for(int k = 1;; k++) {
                if(t.first + t.second * k > fibo[i + 2]) break;
                if(fibo[i] <= t.second) prec[i].push_back(pll(t.second, t.first + t.second * k));
            }
        }
    }

    scanf("%d", &TC);
    while(TC--) main2();
}