#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <bitset>
#include <string>
#include <cmath>
#include <set>
#include <map>
#define Rep(i, x, y) for (int i = x; i <= y; i ++)
#define Dwn(i, x, y) for (int i = x; i >= y; i --)
#define RepE(i, x) for (int i = pos[x]; i; i = g[i].nex)
#define ev g[i].y
#define pll pair<ll, ll>
#define M (1ll << 60)
#define fr first
#define sd second
using namespace std;
typedef long long ll;
typedef double db;
const int N = 300;
const int mod = 1000000007;
int Q, k, s[N];
ll X, Y, ans, f[N];
pair<ll, ll> p[N][N];
void prework() {
	f[0] = 1, f[1] = 1, f[2] = 2, f[3] = 3;
	p[1][1] = pll(1, 2);
	p[1][2] = pll(1, 3);
	s[1] = 2;
	k = 1;
	while (f[k + 1] <= M) {
		k ++;
		f[k + 2] = f[k] + f[k + 1];
		Rep(i, 1, s[k - 1]) {
			ll x = p[k - 1][i].fr, y = p[k - 1][i].sd;
			for (x += y; x <= f[k + 2]; x += y) {
				p[k][++ s[k]] = pll(y, x);
			}
		}
		// cout << s[k]<<endl;
	}
}
void calc(ll x, ll y) {
	if (x <= X && y <= Y && x * 2 > y) (ans += (Y - y) / x + 1) %= mod;
}
int main()
{
	prework();
	scanf ("%d", &Q);
	while(Q --) {
		scanf ("%lld%lld", &X, &Y);
		if (X > Y) swap(X, Y);
		int i = k;
		while (i && p[i][1].fr > X || p[i][1].sd > Y) i --;
		if (!i) i ++;
		if (i == 1) {
			printf("%d %lld\n", i, X % mod * Y % mod);
			continue ;
		}
		ans = 0;
		Rep(j, 1, s[i]) {
			ll x = p[i][j].fr, y = p[i][j].sd;
			calc(x, y);
			calc(y, x);
		}
		printf("%d %lld\n", i, ans);
	}

	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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 <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <bitset>
#include <string>
#include <cmath>
#include <set>
#include <map>
#define Rep(i, x, y) for (int i = x; i <= y; i ++)
#define Dwn(i, x, y) for (int i = x; i >= y; i --)
#define RepE(i, x) for (int i = pos[x]; i; i = g[i].nex)
#define ev g[i].y
#define pll pair<ll, ll>
#define M (1ll << 60)
#define fr first
#define sd second
using namespace std;
typedef long long ll;
typedef double db;
const int N = 300;
const int mod = 1000000007;
int Q, k, s[N];
ll X, Y, ans, f[N];
pair<ll, ll> p[N][N];
void prework() {
	f[0] = 1, f[1] = 1, f[2] = 2, f[3] = 3;
	p[1][1] = pll(1, 2);
	p[1][2] = pll(1, 3);
	s[1] = 2;
	k = 1;
	while (f[k + 1] <= M) {
		k ++;
		f[k + 2] = f[k] + f[k + 1];
		Rep(i, 1, s[k - 1]) {
			ll x = p[k - 1][i].fr, y = p[k - 1][i].sd;
			for (x += y; x <= f[k + 2]; x += y) {
				p[k][++ s[k]] = pll(y, x);
			}
		}
		// cout << s[k]<<endl;
	}
}
void calc(ll x, ll y) {
	if (x <= X && y <= Y && x * 2 > y) (ans += (Y - y) / x + 1) %= mod;
}
int main()
{
	prework();
	scanf ("%d", &Q);
	while(Q --) {
		scanf ("%lld%lld", &X, &Y);
		if (X > Y) swap(X, Y);
		int i = k;
		while (i && p[i][1].fr > X || p[i][1].sd > Y) i --;
		if (!i) i ++;
		if (i == 1) {
			printf("%d %lld\n", i, X % mod * Y % mod);
			continue ;
		}
		ans = 0;
		Rep(j, 1, s[i]) {
			ll x = p[i][j].fr, y = p[i][j].sd;
			calc(x, y);
			calc(y, x);
		}
		printf("%d %lld\n", i, ans);
	}

	return 0;
}
```