#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(ll i=0;i<n;i++)
#define repl(i,l,r) for(ll i=(l);i<(r);i++)
#define per(i,n) for(ll i=(n)-1;i>=0;i--)
#define perl(i,r,l) for(ll i=r-1;i>=l;i--)
#define fi first
#define se second
#define pb push_back
#define ins insert
#define pqueue(x) priority_queue<x,vector<x>,greater<x>>
#define all(x) (x).begin(),(x).end()
#define CST(x) cout<<fixed<<setprecision(x)
#define vtpl(x,y,z) vector<tuple<x,y,z>>
#define rev(x) reverse(x);
using ll=long long;
using vl=vector<ll>;
using vvl=vector<vector<ll>>;
using pl=pair<ll,ll>;
using vpl=vector<pl>;
using vvpl=vector<vpl>;
const ll MOD=1000000007;
const ll MOD9=998244353;
const int inf=1e9+10;
const ll INF=4e18;
const ll dy[9]={0,1,-1,0,1,1,-1,-1,0};
const ll dx[9]={1,0,0,-1,1,-1,1,-1,0};
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
} 

const int mod = 1000000007;
const int max_n = 200005;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }
  bool operator==(const mint &p) const { return x == p.x; }
  bool operator!=(const mint &p) const { return x != p.x; }
  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
using vm=vector<mint>;
using vvm=vector<vm>;
ll P=800;
mint solve(string s){
    while(s.size()<20)s="0"+s;
    string t;
    while(t.size()<10){
        t+=s.back();
        s.pop_back();
    }
    rev(all(t));
    ll n=s.size();
    mint dp[n+1][3][2][2];//mod3,3を含むか,smaller
    rep(i,n+1)rep(j,3)rep(k,2)rep(l,2)dp[i][j][k][l]=0;
    dp[0][0][0][0]=1;
    rep(i,n){
        ll p=s[i]-'0';
        rep(j,3){
            rep(k,10){
                if(k==3){
                    rep(l,2)dp[i+1][(j+k)%3][1][1]+=dp[i][j][l][1];
                }
                else {
                    rep(l,2)dp[i+1][(j+k)%3][l][1]+=dp[i][j][l][1];
                }
            }
            rep(k,p){
                if(k==3){
                    rep(l,2)dp[i+1][(j+k)%3][1][1]+=dp[i][j][l][0];
                }
                else {
                    rep(l,2)dp[i+1][(j+k)%3][l][1]+=dp[i][j][l][0];
                }
            }
            if(p==3){
                rep(l,2)dp[i+1][(j+p)%3][1][0]+=dp[i][j][l][0];
            }
            else{
                rep(l,2)dp[i+1][(j+p)%3][l][0]+=dp[i][j][l][0];
            }
        }
    }
    /*mint ans=0;
    rep(i,3)rep(j,2)rep(k,2){
        cout << dp[n][i][j][k] << endl;
        if(i==0||j==1)ans+=dp[n][i][j][k];
    }
    return ans;*/
    ll m=t.size();
    mint ndp[m+1][3][2][2][P];//さらにmodpを加える
    rep(i,m+1)rep(j,3)rep(k,2)rep(l,2)rep(t,P)ndp[i][j][k][l][t]=0;
    rep(i,3)rep(j,2)rep(k,2)ndp[0][i][j][k][0]+=dp[n][i][j][k];
    rep(i,m){
        ll p=t[i]-'0';
        rep(j,3){
            rep(k,P){
                rep(to,10){
                    if(to==3){
                        rep(l,2)ndp[i+1][(j+to)%3][1][1][(k*10+to)%P]+=ndp[i][j][l][1][k];
                    }
                    else{
                        rep(l,2)ndp[i+1][(j+to)%3][l][1][(k*10+to)%P]+=ndp[i][j][l][1][k];
                    }
                }
                rep(to,p){
                    if(to==3){
                        rep(l,2)ndp[i+1][(j+to)%3][1][1][(k*10+to)%P]+=ndp[i][j][l][0][k];
                    }
                    else{
                        rep(l,2)ndp[i+1][(j+to)%3][l][1][(k*10+to)%P]+=ndp[i][j][l][0][k];
                    }
                }
                if(p==3){
                    rep(l,2)ndp[i+1][(j+p)%3][1][0][(k*10+p)%P]+=ndp[i][j][l][0][k];
                }
                else{
                    rep(l,2)ndp[i+1][(j+p)%3][l][0][(k*10+p)%P]+=ndp[i][j][l][0][k];
                }
            }
        }
    }
    mint ans=0;
    rep(j,3)rep(k,2)rep(l,2)rep(t,P){
        if((t!=0)&&(j==0||k==1))ans+=ndp[m][j][k][l][t];
    }
    return ans;
}
int main(){
    string a,b;cin >> a >> b;
    cin >> P;
    mint ans=solve(b)-solve(a);
    ll th=0,ex=0,ei=0;
    rep(i,a.size()){
        th*=10;ei*=10;
        ll p=a[i]-'0';
        if(p==3)ex=1;
        th+=p;
        ei+=p;
        th%=3;
        ei%=P;
    }
    if(((th==0)||ex)&&(ei!=0)){
        ans+=1;
    }
    cout << ans << endl;
}