/*CodeForces 55D Beautiful numbers(数位dp)problem:问[l,r]之间有多少个能被它所包含的非零数整除solve:如果能这些数整除,则能被他们的最小公倍数整除. 1~9的最小公倍数为2520,所以在过程中维护lcm和这个数对2520的取余而且可以处理出1~9的所有lcm,离散化处理.hhh-2016-08-25 16:56:23*/#pragma comment(linker,"/STACK:124000000,124000000")#include <algorithm>#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <vector>#include <math.h>#include <queue>#include <map>#define lson  i<<1#define rson  i<<1|1#define ll long long#define clr(a,b) memset(a,b,sizeof(a))#define scanfi(a) scanf("%d",&a)#define scanfl(a) scanf("%I64d",&a)#define key_val ch[ch[root][1]][0]#define inf 0x3f3f3f3fusing namespace std;const ll mod = 2520;const int maxn = 100010;int tot,t[maxn];ll dp[20][3000][50];int bin[3000];int cnt = 0;int lcm(int a,int b){    if(!a && !b)        return 0;    if(!a)  return b;    if(!b)  return a;    int ta=a,tb=b;    while(a % b != 0)    {        int p = a % b;        a = b;        b = p;    }    return ta/b*tb;}int fin(int pos){    int l = 0,r = cnt-1;    while(l < r)    {        int mid = (l+r) >> 1;        if(bin[mid] > pos)        {            r = mid -1;        }        else if(bin[mid] == pos)        {            return mid;        }        else            l = mid + 1;    }}ll dfs(int len,int nex,int div,int flag){    if(len < 0 && div == 0)        return 0;    if(len < 0)        return nex % div == 0;    int tp = bin[div];    if(dp[len][nex][tp] != -1 && !flag)        return dp[len][nex][tp] ;    ll ans = 0;    int n = flag ? t[len] : 9;    for(int i = 0; i <= n; i++)    {        int lc;        lc = lcm(div,i);//        cout << lc <<" " << i <<endl;        int ta = (nex*10+i) % mod;        ans += dfs(len-1,ta,lc,flag && i == n);    }//    cout << ans<<" "<<div<<endl;    if(!flag)        dp[len][nex][tp] = ans;    return ans;}ll cal(ll a){    tot = 0;    while(a)    {        int p = a%10LL;        t[tot++] = p;        a /= 10LL;    }    ll ans = 0;    ans += dfs(tot-1,0,0,1);    return ans;}ll a,b;int main(){    int T;    cnt = 0;    bin[cnt++] = 0;    for(int i = 1;i <= mod;i++)    {        if(mod % i == 0)           bin[i] = cnt++;    }//    freopen("in.txt","r",stdin);    memset(dp,-1,sizeof(dp));    scanfi(T);    while(T--)    {        scanfl(a),scanfl(b);        printf("%I64d\n",cal(b) - cal(a-1));    }}