#include <bits/stdc++.h>
using namespace std;
const int N = 100;
const int MOD = 1e9 + 7;
#define LL long long
#define pI pair <LL, LL>
#define fr first
#define sc second
#define mp make_pair
LL f[N]; int tot;
vector <pI> S[N];
int q;

void print(vector <pI> a)
{
    for (int i = 0; i < a.size(); ++ i)
        cerr << a[i].fr << " " << a[i].sc << endl;
}
int main()
{
    f[0] = f[1] = 1; tot = 1;
    while (f[tot] + f[tot - 1] <= 5e18)
    {
        tot ++;
        f[tot] = f[tot - 1] + f[tot - 2];
    }
    S[1].push_back(mp(1, 2));
    S[1].push_back(mp(1, 3));
    S[1].push_back(mp(1, 4));
    for (int i = 2; i <= tot - 3; ++ i)
        for (int j = 0; j < S[i - 1].size(); ++ j)
            for (int k = 1; k * S[i - 1][j].sc + S[i - 1][j].fr <= f[i + 2]; ++ k)
                S[i].push_back(mp(S[i - 1][j].sc, k * S[i - 1][j].sc + S[i - 1][j].fr));
    scanf("%d", &q);
    while (q --)
    {
        LL x, y, ans = 0;
        scanf("%lld%lld", &x, &y);
        if (x > y) swap(x, y);
        int k = 1;
        while (f[k + 1] <= x && f[k + 2] <= y) k ++;
        for (int i = 0; i < S[k - 1].size(); ++ i)
            if (S[k - 1][i].sc <= x)
                ans += (y - S[k - 1][i].fr) / S[k - 1][i].sc;

        for (int i = 0; i < S[k - 1].size(); ++ i)
            if (S[k - 1][i].sc <= y)
                ans += (x - S[k - 1][i].fr) / S[k - 1][i].sc;
        if (x == 1) ans = y;
        if (x == 2 && y == 2) ans = 4;
        printf("%d %d\n", k, (int)(ans % MOD));
    }
}

### Prompt

Generate a Cpp solution 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;
const int N = 100;
const int MOD = 1e9 + 7;
#define LL long long
#define pI pair <LL, LL>
#define fr first
#define sc second
#define mp make_pair
LL f[N]; int tot;
vector <pI> S[N];
int q;

void print(vector <pI> a)
{
    for (int i = 0; i < a.size(); ++ i)
        cerr << a[i].fr << " " << a[i].sc << endl;
}
int main()
{
    f[0] = f[1] = 1; tot = 1;
    while (f[tot] + f[tot - 1] <= 5e18)
    {
        tot ++;
        f[tot] = f[tot - 1] + f[tot - 2];
    }
    S[1].push_back(mp(1, 2));
    S[1].push_back(mp(1, 3));
    S[1].push_back(mp(1, 4));
    for (int i = 2; i <= tot - 3; ++ i)
        for (int j = 0; j < S[i - 1].size(); ++ j)
            for (int k = 1; k * S[i - 1][j].sc + S[i - 1][j].fr <= f[i + 2]; ++ k)
                S[i].push_back(mp(S[i - 1][j].sc, k * S[i - 1][j].sc + S[i - 1][j].fr));
    scanf("%d", &q);
    while (q --)
    {
        LL x, y, ans = 0;
        scanf("%lld%lld", &x, &y);
        if (x > y) swap(x, y);
        int k = 1;
        while (f[k + 1] <= x && f[k + 2] <= y) k ++;
        for (int i = 0; i < S[k - 1].size(); ++ i)
            if (S[k - 1][i].sc <= x)
                ans += (y - S[k - 1][i].fr) / S[k - 1][i].sc;

        for (int i = 0; i < S[k - 1].size(); ++ i)
            if (S[k - 1][i].sc <= y)
                ans += (x - S[k - 1][i].fr) / S[k - 1][i].sc;
        if (x == 1) ans = y;
        if (x == 2 && y == 2) ans = 4;
        printf("%d %d\n", k, (int)(ans % MOD));
    }
}

```