#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <set>
#include <map>
#include <queue>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <cctype>
#include <cassert>
#include <limits>
#include <functional>
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
#if defined(_MSC_VER) || __cplusplus > 199711L
#define aut(r,v) auto r = (v)
#else
#define aut(r,v) __typeof(v) r = (v)
#endif
#define each(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it)
#define all(o) (o).begin(), (o).end()
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset(m,v,sizeof(m))
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3fLL
using namespace std;
typedef vector<int> vi; typedef pair<int,int> pii; typedef vector<pair<int,int> > vpii; typedef long long ll;
template<typename T, typename U> inline void amin(T &x, U y) { if(y < x) x = y; }
template<typename T, typename U> inline void amax(T &x, U y) { if(x < y) x = y; }


template<int MOD>
struct ModInt {
	static const int Mod = MOD;
	unsigned x;
	ModInt() : x(0) {}
	ModInt(signed sig) { int sigt = sig % MOD; if(sigt < 0) sigt += MOD; x = sigt; }
	ModInt(signed long long sig) { int sigt = sig % MOD; if(sigt < 0) sigt += MOD; x = sigt; }
	int get() const { return (int)x; }

	ModInt &operator+=(ModInt that) { if((x += that.x) >= MOD) x -= MOD; return *this; }
	ModInt &operator-=(ModInt that) { if((x += MOD - that.x) >= MOD) x -= MOD; return *this; }
	ModInt &operator*=(ModInt that) { x = (unsigned long long)x * that.x % MOD; return *this; }

	ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
	ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
	ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
};
typedef ModInt<1000000007> mint;

vector<int> digits;
int P;
vector<mint> memo;
bool eqOK;
mint rec(int i, int m, bool three, bool lt) {
	mint &r = memo[((i + 1) * 3 + m) * 4 + three * 2 + lt];
	if(r.x != -1) return r;
	if(i <= 4) {
		r = mint();
		int p = 1;
		int es = 0;
		rep(j, i + 1) {
			es += p * digits[j];
			p *= 10;
		}
		rep(ds, p) if(lt || ds <= es) {
			bool nlt = lt || ds < es;
			int nm = (m * p + ds) % 3;
			bool nthree = three;
			int t = ds;
			rep(j, i+1) {
				nthree |= t % 10 == 3;
				t /= 10;
			}
			bool ok = true;
			ok &= nlt || eqOK;
			ok &= nm == 0 || nthree;
			ok &= ds % P != 0;
			if(ok) r += 1;
		}
		return r;
	}
	r = mint();
	int e = digits[i];
	rep(d, 10) if(lt || d <= e)
		r += rec(i-1, (m * 10 + d) % 3, three || d == 3, lt || d < e);
	return r;
}
mint solve(const char *A) {
	int len = strlen(A);
	digits.resize(len);
	rep(i, len)
		digits[i] = A[len - 1 - i] - '0';
	mint r;
	mint undef; undef.x = -1;
	memo.assign((len+1) * 3 * 4, undef);
	return rec(len - 1, 0, false, false);
}

int main() {
	char *A = new char[200002], *B = new char[200002];
	while(~scanf("%s%s%d", A, B, &P)) {
		mint ans;
		eqOK = true;
		ans += solve(B);
		eqOK = false;
		ans -= solve(A);
		printf("%d\n", ans.get());
	}
	return 0;
}