Problem: Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. We will not argue with this and just count the quantity of beautiful numbers in given ranges.

Input

The first line of the input contains the number of cases t (1 ≤ t ≤ 10). Each of the next t lines contains two natural numbers li and ri (1 ≤ li ≤ ri ≤ 9 ·1018).

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).

Output

Output should contain t numbers — answers to the queries, one number per line — quantities of beautiful numbers in given intervals (from li to ri, inclusively).

Examples

Input

1
1 9


Output

9


Input

1
12 15


Output

2

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class D {

	final static int[] a = new int[] { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14,
			15, 18, 20, 21, 24, 28, 30, 35, 36, 40, 42, 45, 56, 60, 63, 70, 72,
			84, 90, 105, 120, 126, 140, 168, 180, 210, 252, 280, 315, 360, 420,
			504, 630, 840, 1260, 2520 };

	long[][][] d;
	long[] pows;
	final static int MAXLEN = 20;
	final static int MOD = 2520;

	int gcd(int a, int b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}

	int lcm(int a, int b) {
		if (b == 0 || a == 0)
			return a | b;
		return a / gcd(a, b) * b;
	}

	void precalc() {
		pows = new long[MAXLEN];
		pows[0] = 1;
		for (int i = 1; i < pows.length; i++) {
			pows[i] = pows[i - 1] * 10;
		}
		d = new long[MAXLEN][a.length][MOD];
		d[0][0][0] = 1;
		int pow = 1;
		for (int len = 0; len < MAXLEN - 1; len++) {
			for (int it = 0; it < a.length; it++) {
				int lcm0 = a[it];
				for (int digit = 0; digit < 10; digit++) {
					int lcm1 = lcm(digit, lcm0);
					int ind = Arrays.binarySearch(a, lcm1);
					int r1 = digit * pow % MOD;
					for (int r = 0; r < MOD; r++) {
						d[len + 1][ind][r1] += d[len][it][r];
						r1++;
						if (r1 == MOD)
							r1 = 0;
					}
				}
			}
			pow = (pow * 10) % MOD;
		}
	}

	void solve() throws IOException {
		precalc();
		long t = nextLong();
		for (int i = 0; i < t; i++) {
			long l = nextLong(), r = nextLong();
			System.out.println(get(r) - get(l - 1));
		}
	}

	long get(long max) {
		if (max == 0)
			return 0;
		String s = Long.toString(max);
		int len = s.length();

		long res = -1;
		int lcm = 1;
		for (int it = 0; it < len; it++) {
			long n1 = it == 0 ? 0 : max - (max % pows[len - it]);
			int curdigit = s.charAt(it) - '0';
			for (int digit = 0; digit < curdigit; digit++) {
				long n2 = n1 + digit * pows[len - it - 1];
				int lcm1 = lcm(lcm, digit);
				int r = (int) ((MOD - n2 % MOD) % MOD);

				for (int i = 0; i < a.length; i++) {
					int l = lcm(lcm1, a[i]);
					for (int j = r % l; j < MOD; j += l) {
						res += d[len - it - 1][i][j];
					}
				}
			}
			lcm = lcm(lcm, curdigit);
		}
		if (max % lcm == 0)
			res++;
		return res;
	}

	public static void main(String[] args) {
		try {
			new D().solve();
		} catch (IOException e) {
			e.printStackTrace();
			System.exit(239);
		}
	}

	static long nextLong() throws IOException {
		long m = 0;
		int c = System.in.read();
		while (c < '0' || c > '9')
			c = System.in.read();
		while (c >= '0' && c <= '9') {
			m = m * 10 + (c - '0');
			c = System.in.read();
		}
		return m;
	}
}