#include <queue>
#include <vector>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MP make_pair
#define ll long long
#define fi first
#define se second
using namespace std;

template <typename T>
void read(T &x) {
    x = 0; bool f = 0;
    char c = getchar();
    for (;!isdigit(c);c=getchar()) if (c=='-') f=1;
    for (;isdigit(c);c=getchar()) x=x*10+(c^48);
    if (f) x=-x;
}

template <typename T>
inline void Mx(T &x, T y) { x < y && (x = y); }

template <typename T>
inline void Mn(T &x, T y) { x > y && (x = y); }

const int P = 1e9+7;
const ll INF = 1e18;
ll fib[105], l;
ll m, T, n;


vector<pair<ll, ll> > res[100];
int main() {
	for (fib[0] = fib[1] = l = 1; fib[l] <= INF; l++)
		fib[l+1] = fib[l] + fib[l-1];
		
	res[1].push_back(MP(1, 2));
	res[1].push_back(MP(1, 3));
	res[1].push_back(MP(1, 4));
	for (int i = 1;i < l - 2; i++) {
		for (auto j: res[i]) {
			ll x = j.fi, y = j.se;
			x += y;
			while (x <= fib[i+3] + fib[i]) {
				res[i+1].push_back(MP(y, x));
				x += y;
			}
		}
	}
	
	for (read(T); T; T--) {
		read(n), read(m);
		n > m && (swap(n, m), 0);
		int k = 1;
		while (fib[k+1] <= n && fib[k+2] <= m) ++k;
		printf ("%d ", k);
		if (k == 1) {
			printf ("%lld\n", n * m % P);
			continue;
		}
		ll ans = 0;
		for (auto j: res[k-1]) {
			ll x = j.fi, y = j.se;
			if (y <= n) ans += (m - x) / y;
			if (y <= m) ans += (n - x) / y;
			ans %= P;
		}
		printf ("%lld\n", ans);
	}
	return 0;
}
### Prompt

Please formulate 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 <queue>
#include <vector>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MP make_pair
#define ll long long
#define fi first
#define se second
using namespace std;

template <typename T>
void read(T &x) {
    x = 0; bool f = 0;
    char c = getchar();
    for (;!isdigit(c);c=getchar()) if (c=='-') f=1;
    for (;isdigit(c);c=getchar()) x=x*10+(c^48);
    if (f) x=-x;
}

template <typename T>
inline void Mx(T &x, T y) { x < y && (x = y); }

template <typename T>
inline void Mn(T &x, T y) { x > y && (x = y); }

const int P = 1e9+7;
const ll INF = 1e18;
ll fib[105], l;
ll m, T, n;


vector<pair<ll, ll> > res[100];
int main() {
	for (fib[0] = fib[1] = l = 1; fib[l] <= INF; l++)
		fib[l+1] = fib[l] + fib[l-1];
		
	res[1].push_back(MP(1, 2));
	res[1].push_back(MP(1, 3));
	res[1].push_back(MP(1, 4));
	for (int i = 1;i < l - 2; i++) {
		for (auto j: res[i]) {
			ll x = j.fi, y = j.se;
			x += y;
			while (x <= fib[i+3] + fib[i]) {
				res[i+1].push_back(MP(y, x));
				x += y;
			}
		}
	}
	
	for (read(T); T; T--) {
		read(n), read(m);
		n > m && (swap(n, m), 0);
		int k = 1;
		while (fib[k+1] <= n && fib[k+2] <= m) ++k;
		printf ("%d ", k);
		if (k == 1) {
			printf ("%lld\n", n * m % P);
			continue;
		}
		ll ans = 0;
		for (auto j: res[k-1]) {
			ll x = j.fi, y = j.se;
			if (y <= n) ans += (m - x) / y;
			if (y <= m) ans += (n - x) / y;
			ans %= P;
		}
		printf ("%lld\n", ans);
	}
	return 0;
}
```