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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Map.Entry;
import java.util.StringTokenizer;
import java.util.TreeMap;

public class Div2_421E {

	static int[] numPF = new int[1_000_001];
	static int[][] pF = new int[1_000_001][7];
	static int[][] pFC = new int[1_000_001][7];

	static void sieve() {
		for (int cInt = 2; cInt <= 1_000_000; cInt++) {
			if (numPF[cInt] == 0) {

				for (int mult = cInt; mult <= 1_000_000; mult += cInt) {
					int pfI = numPF[mult]++;
					pF[mult][pfI] = cInt;
					pFC[mult][pfI] = 1;
				}

				for (long pow = (long) cInt * cInt; pow <= 1_000_000; pow *= cInt) {
					int powI = (int) pow;
					for (int mult = powI; mult <= 1_000_000; mult += powI) {
						pFC[mult][numPF[mult] - 1]++;
					}
				}
			}
		}
	}

	static int sNumPF;
	static int[] sPF;
	static int[] sPFC;

	public static long compute(int[] nF, int[] mF, int[] sF) {
		long n = (long) nF[0] * nF[1] * nF[2];
		long m = (long) mF[0] * mF[1] * mF[2];

		TreeMap<Integer, Integer> nPFMap = new TreeMap<Integer, Integer>();
		TreeMap<Integer, Integer> sPFMap = new TreeMap<Integer, Integer>();

		for (int i = 0; i < 3; i++) {
			int cF = nF[i];
			for (int j = 0; j < numPF[cF]; j++) {
				int cPrime = pF[cF][j];
				Integer cCnt = nPFMap.get(cPrime);
				if (cCnt == null) {
					nPFMap.put(cPrime, pFC[cF][j]);
				} else {
					nPFMap.put(cPrime, cCnt + pFC[cF][j]);
				}
			}
			cF = sF[i];
			for (int j = 0; j < numPF[cF]; j++) {
				int cPrime = pF[cF][j];
				Integer cCnt = sPFMap.get(cPrime);
				if (cCnt == null) {
					sPFMap.put(cPrime, pFC[cF][j]);
				} else {
					sPFMap.put(cPrime, cCnt + pFC[cF][j]);
				}
			}
		}

		if (sPFMap.containsKey(2)) {
			sPFMap.put(2, sPFMap.get(2) + 1);
		} else {
			sPFMap.put(2, 1);
		}

		// from here on, everything that refers to s is actually 2s

		int nNumPF = 0;
		sNumPF = 0;

		int[] nPF = new int[15];
		sPF = new int[15];
		int[] nPFC = new int[15];
		sPFC = new int[15];

		for (Entry<Integer, Integer> cEntry : nPFMap.entrySet()) {
			nPF[nNumPF] = cEntry.getKey();
			nPFC[nNumPF++] = cEntry.getValue();
		}
		for (Entry<Integer, Integer> cEntry : sPFMap.entrySet()) {
			sPF[sNumPF] = cEntry.getKey();
			sPFC[sNumPF++] = cEntry.getValue();
		}

		int nInd = 0;
		int sInd = 0;

		int nFbdn = 0;
		long[] fbdn = new long[15];

		while (nInd < nNumPF && sInd < sNumPF) {
			if (nPF[nInd] < sPF[sInd]) {
				fbdn[nFbdn++] = nPF[nInd];
				nInd++;
			} else if (nPF[nInd] == sPF[sInd]) {
				if (nPFC[nInd] > sPFC[sInd]) {
					fbdn[nFbdn++] = pow(sPF[sInd], sPFC[sInd] + 1);
				}
				nInd++;
				sInd++;
			} else {
				sInd++;
			}
		}
		
		while (nInd < nNumPF){
			fbdn[nFbdn++] = nPF[nInd];
			nInd++;
		}

		// PIE to count number of integers which are not divisible by any of fbdn
		// can be optimized more

		long cnt = m;
		for (int bSet = 1; bSet < (1 << nFbdn); bSet++) {
			long prod = 1;

			int copy = bSet;
			int cInd = 0;
			int pol = 0;
			while (copy > 0) {
				if ((copy & 1) == 1) {
					prod *= fbdn[cInd];
					pol ^= 1;
				}
				cInd++;
				copy >>= 1;
			}

			if (pol == 1) {
				cnt -= m / prod;
			} else {
				cnt += m / prod;
			}
		}
		
		long phase1 = cnt;

		long phase2 = cntLess(1, 0, n);

		return phase1 + phase2;
	}
	
	static long pow(long base, int exp) {
		long cur = 1;
		while (exp-- > 0) {
			cur *= base;
		}
		return cur;
	}

	static long cntLess(long cProd, int cPI, long bound) {
		if (cProd > bound) {
			return 0;
		}

		if (cPI == sNumPF) {
			return 1;
		}

		long nProd = cProd;
		long sum = 0;
		for (int numU = 0; numU <= sPFC[cPI]; numU++) {
			sum += cntLess(nProd, cPI + 1, bound);
			nProd *= sPF[cPI];
		}
		return sum;
	}

	public static void main(String[] args) throws IOException {
		sieve();
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter printer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		int numT = Integer.parseInt(reader.readLine());

		while (numT-- > 0) {
			StringTokenizer inputData = new StringTokenizer(reader.readLine());
			int[] n = new int[] { Integer.parseInt(inputData.nextToken()), Integer.parseInt(inputData.nextToken()),
					Integer.parseInt(inputData.nextToken()) };
			inputData = new StringTokenizer(reader.readLine());
			int[] m = new int[] { Integer.parseInt(inputData.nextToken()), Integer.parseInt(inputData.nextToken()),
					Integer.parseInt(inputData.nextToken()) };
			inputData = new StringTokenizer(reader.readLine());
			int[] s = new int[] { Integer.parseInt(inputData.nextToken()), Integer.parseInt(inputData.nextToken()),
					Integer.parseInt(inputData.nextToken()) };
			printer.println(compute(n, m, s));
		}
		printer.close();
	}

}