#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#define lo long long
#define mod 1000000007
using namespace std;
int a[23];
lo mi[41];        //mi[i]:10的i次幂
inline lo read()
{
	register long long ans=0,f=1;char ch=getchar();
	while(!isdigit(ch)) {if(ch==‘-‘) f=-1;ch=getchar();}
	while(isdigit(ch)) {ans=ans*10+ch-‘0‘;ch=getchar();}
	return ans*f;
}
struct num{
	lo su,sq,nu;        //和，平方和，个数 
	num(lo u=0,lo q=0,lo n=0):su(u),sq(q),nu(n){}
}dp[23][11][11];
num dfs(int wi,int o,lo e,bool lim)      //哪一位，每一位%7，数%7，限制？ 
{
	num ans,zc;
	if(wi<1)
	{
		ans.nu=o&&e;
		return ans;
	}
	if(!lim&&dp[wi][o][e].su>-1)
	  return dp[wi][o][e];
	int l=lim? a[wi]:9;
	for(lo i=0;i<=l;i++)
	  if(i!=7)       //////
	  {
	  	 zc=dfs(wi-1,(o+i)%7,(lo)(e*10+i)%7,lim&&i==a[wi]);          //e+i*mi[wi-1]?
	  	 ans.sq=(((ans.sq+i*i*mi[2*wi-2]%mod*zc.nu%mod)%mod+zc.sq)%mod+(lo)2*zc.su*i%mod*mi[wi-1]%mod)%mod;
	  	 ans.su=((ans.su+zc.su)%mod+zc.nu*i%mod*mi[wi-1]%mod)%mod;
	  	 ans.nu=(ans.nu+zc.nu)%mod;
	  }
	if(!lim)
	  dp[wi][o][e]=ans;
	return ans;
}
lo sol(lo x)
{
	int w=0;
	while(x)
	{
		a[++w]=x%10;
		x/=10;
	}
	num ans=dfs(w,0,(lo)0,1);
	return ans.sq;
}
int main()
{
	int t;lo l,r;
	t=read();
	mi[0]=1;
	for(int i=1;i<=38;i++)
	  mi[i]=mi[i-1]*10%mod;
	memset(dp,-1,sizeof(dp));
	while(t--)
	{
		l=read();r=read();
		lo rr=sol(r);
		lo ll=sol(l-1);
		printf("%lld\n",(rr+mod-ll)%mod);
	}
	return 0;
}