#include "bits/stdc++.h"
using namespace std;

static const int MOD = 1000000007;

long long fib[100];
vector<pair<long long, long long>> p[100];

int main() {
        fib[0] = fib[1] = 1;
        for (int k = 2; k <= 90; k ++) {
                fib[k] = fib[k - 1] + fib[k - 2];
                for (int j = 2; j <= k + 1; j ++) {
                        long long a = 1, b = 1;
                        for (int x = 2; x <= k; x ++) {
                                long long c = a + b;
                                if (x == j) c += a;
                                b = a;
                                a = c;
                        }
                        p[k].emplace_back(a, a + b);
                        p[k].emplace_back(a + b, a);
                }
        }
        int q;
        cin >> q;
        while (q --) {
                long long x, y;
                cin >> x >> y;
                if (x > y) swap(x, y);
                if ((x == 1) || (x == 2 && y == 2)) {
                        cout << 1 << ' ' << x * y % MOD << endl;
                        continue;
                }
                int k = 0;
                while (fib[k + 1] <= x && fib[k + 2] <= y) k ++;
                long long ans = 0;
                for (auto pp : p[k]) if (x >= pp.first && y >= pp.second) {
                        ans = (ans + 1 + (y - pp.second) / pp.first) % MOD;
                }
                cout << k << ' ' << ans << endl;
        }
        return 0;
}


### Prompt

Please create a solution in Cpp to the following 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

### Response

```cpp
#include "bits/stdc++.h"
using namespace std;

static const int MOD = 1000000007;

long long fib[100];
vector<pair<long long, long long>> p[100];

int main() {
        fib[0] = fib[1] = 1;
        for (int k = 2; k <= 90; k ++) {
                fib[k] = fib[k - 1] + fib[k - 2];
                for (int j = 2; j <= k + 1; j ++) {
                        long long a = 1, b = 1;
                        for (int x = 2; x <= k; x ++) {
                                long long c = a + b;
                                if (x == j) c += a;
                                b = a;
                                a = c;
                        }
                        p[k].emplace_back(a, a + b);
                        p[k].emplace_back(a + b, a);
                }
        }
        int q;
        cin >> q;
        while (q --) {
                long long x, y;
                cin >> x >> y;
                if (x > y) swap(x, y);
                if ((x == 1) || (x == 2 && y == 2)) {
                        cout << 1 << ' ' << x * y % MOD << endl;
                        continue;
                }
                int k = 0;
                while (fib[k + 1] <= x && fib[k + 2] <= y) k ++;
                long long ans = 0;
                for (auto pp : p[k]) if (x >= pp.first && y >= pp.second) {
                        ans = (ans + 1 + (y - pp.second) / pp.first) % MOD;
                }
                cout << k << ' ' << ans << endl;
        }
        return 0;
}


```