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>
#define LL long long
using namespace std;

const int MOD = 1000000007;
const int LOG = 100;

vector<pair<LL, LL> > num[LOG];

inline LL read() {
	char c = getchar(); LL ret = 0, f = 1;
	for (; c < '0' || c > '9'; f = c == '-'? -1: 1, c = getchar());
	for (; '0' <= c && c <= '9'; ret = ret * 10 + c - '0', c = getchar());
	return ret * f;
}

inline void solve(LL A, LL B) {
	if (A < num[2][0].first || B < num[2][0].second) {
		printf("1 %d\n", A * B % MOD);
		return;
	}
	for (int i = 2; i < LOG; i++) {
		//cout<<i<<' '<<A<<' '<<num[i + 1][0].first<<' '<<B<<' '<<num[i + 1][0].second<<endl;
		if (A < num[i + 1][0].first || B < num[i + 1][0].second) {
			int cnt = 0;
			for (int j = 0; j < (int)num[i].size(); j++) {
				LL a = num[i][j].first, b = num[i][j].second;
				cnt = (cnt + (a <= A && b <= B? (B - b) / a + 1: 0)) % MOD;
				cnt = (cnt + (b <= A && a <= B? (A - b) / a + 1: 0)) % MOD;
			}
			printf("%d %d\n", i, cnt);
			return;
		}
	}
}

inline void prework() {
	num[0] = vector<pair<LL,LL> >{make_pair(1, 1)};
	for (int i = 1; i < LOG; i++) {
		for (int j = 0; j < (int)num[i - 1].size(); ++j) {
			LL a = num[i - 1][j].first, b = num[i - 1][j].second;
			num[i].push_back(make_pair(b, a + b));
		}
		LL a = num[i - 1][0].first, b = num[i - 1][0].second;
		num[i].push_back(make_pair(a + b, 2 * a + b));
		//cout<<num[i][0].first<<' '<<num[i][0].second<<endl;
	}
}

int main() {
	//freopen("11input.in", "r", stdin);
	prework();
	for (int T = read(); T; T--) {
		LL A = read(), B = read();
		solve(min(A, B), max(A, B));
	}
	return 0;
}