Problem: Mister B has a house in the middle of a giant plain field, which attracted aliens life. For convenience, aliens specified the Cartesian coordinate system on the field in such a way that Mister B's house has coordinates (0, 0). After that they sent three beacons to the field, but something went wrong. One beacon was completely destroyed, while the other two landed in positions with coordinates (m, 0) and (0, n), respectively, but shut down.

Mister B was interested in this devices, so he decided to take them home. He came to the first beacon, placed at (m, 0), lifted it up and carried the beacon home choosing the shortest path. After that he came to the other beacon, placed at (0, n), and also carried it home choosing the shortest path. When first beacon was lifted up, the navigation system of the beacons was activated.

Partially destroyed navigation system started to work in following way.

At time moments when both survived beacons are at points with integer coordinates the system tries to find a location for the third beacon. It succeeds if and only if there is a point with integer coordinates such that the area of the triangle formed by the two survived beacons and this point is equal to s. In this case the system sends a packet of information with beacon positions to aliens, otherwise it doesn't.

Compute how many packets of information system sent while Mister B was moving the beacons.

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. The next 3·t lines describe t test cases. 

Every test case is described in three lines as follows. Note that each parameter is given as a product of three factors.

The first line of a test case contains three space-separated integers: n1, n2, n3 (1 ≤ ni ≤ 106) such that n = n1·n2·n3.

The second line contains three space-separated integers: m1, m2, m3 (1 ≤ mi ≤ 106) such that m = m1·m2·m3.

The third line contains three space-separated integers: s1, s2, s3 (1 ≤ si ≤ 106) such that s = s1·s2·s3.

Note that for hacks only tests with t = 1 allowed.

Output

Print t integers one per line — the answers for each test.

Example

Input

3
2 1 1
2 1 1
1 1 3
1 5 1
2 2 1
1 1 2
10 6 18
2 103 2
13 1 13


Output

4
7
171

Note

First test case contains the following beacon positions: (2, 0) and (0, 2), s = 3. The following packets could be sent: ((2, 0), (0, 2), ( - 1, 0)), ((1, 0), (0, 2), (4, 0)), ((0, 0), (0, 2), (3, 1)), ((0, 0), (0, 1), ( - 6, 0)), where (b1, b2, p) has next description: b1 — first beacon position, b2 — second beacon position, p — some generated point.

Second test case contains the following beacon initial positions: (4, 0) and (0, 5), s = 2. The following packets could be sent: ((4, 0), (0, 5), (0, 4)), ((3, 0), (0, 5), (2, 3)), ((2, 0), (0, 5), (2, 2)), ((1, 0), (0, 5), (1, 4)), ((0, 0), (0, 4), (0, - 1)), ((0, 0), (0, 2), (2, 0)), ((0, 0), (0, 1), (4, 0)).

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class abc {

	static int MAX = 2_000_002;
	static int[] lpf;
	static boolean isPrime[];
	
	static int[] N, S, M;
	static long eN, eS, eM;
	
	static int[] pfs, primeIndex, version, pcnt;
	static int pfi, V;
	
	static long[] as;
	static int ai;
	
	static long answer;
	
	public static void main(String[] args) {
		FS in = new FS();
		PrintWriter out = new PrintWriter(System.out);
		isPrime = new boolean[MAX];
		lpf = new int[MAX];
		Arrays.fill(isPrime, true);
		isPrime[0] = isPrime[1] = false;
		for(int i = 2; i < MAX; i++) {
			if(isPrime[i]) {
				lpf[i] = i;
				for(int j = i<<1; j < MAX; j+=i) {
					isPrime[j] = false;
					lpf[j] = i;
				}
			}
		}
		
		pfs = new int[30];
		pcnt = new int[30];
		primeIndex = new int[MAX];
		version = new int[MAX];
		V = 14;
		pfi = 0;
		
		int T = in.nextInt();
		for(int runs = 1; runs <= T; runs++) {
			answer = 0;
			N = read(in); eN = expand(N);
			M = read(in); eM = expand(M);
			S = read(in); S[0] *= 2;
			eS = expand(S);
			
			factor(S);
			dfs(0, 1);
			
			factor(N);
			calcA();
			ie(0,0,1);
			out.println(answer);
		}
		
		out.close();
	}
	
	static void ie(int id, int cnt, long val) {
		if(id >= ai) {
			if((cnt&1) == 0) answer += (eM/val);
			else answer -= (eM/val);
		}
		else {
			ie(id+1, cnt+1, val*as[id]);
			ie(id+1, cnt, val);
		}
	}
	
	static void calcA() {
		as = new long[30];
		ai = 0;
		for(int i = 0; i < pfi; i++) {
			long here = pfs[i];
			for(int j = 0; j < pcnt[i]; j++) {
				if(eS % here != 0) {
					as[ai++] = here;
					break;
				}
				here *= pfs[i];
			}
		}
	}
	
	static void dfs(int index, long num) {
		if(num > eN) return;
		if(index >= pfi) { answer++; return;}
		
		long toMult = 1;
		for(int take = 0; take <= pcnt[index]; take++) {
			dfs(index+1, num*toMult);
			toMult *= pfs[index];
		}
	}
	
	static void factor(int[] vals) {
		V++;
		pfi = 0;
		for(int val : vals) {
			while(val != 1) {
				int pf = lpf[val];
				if(version[pf] != V) {
					version[pf] = V;
					pfs[pfi] = pf;
					pcnt[pfi] = 0;
					primeIndex[pf] = pfi;
					pfi++;
				}
				pcnt[primeIndex[pf]]++;
				val /= pf;
			}
		}
	}

	static long expand(int[] x) {
		long res = 1;
		for(int xx : x) res *= xx;
		return res;
	}
	
	static int[] read(FS in){
		int[] res = new int[3];
		for(int i = 0; i < 3; i++) res[i] = in.nextInt();
		return res;
	}
	
	static int[] mult2(int[] x) {
		int[] res = Arrays.copyOf(x, x.length);
		res[0] *= 2;
		return res;
	}
	
	static class FS{
		BufferedReader br;
		StringTokenizer st;
		public FS() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		String next() {
			while(st == null || !st.hasMoreElements()) {
				try {st = new StringTokenizer(br.readLine());}
				catch(Exception e) {}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
	}
}