#include <stdio.h>

int a, b, c, k;
int d[15][15][15];

void calc_din() {
	if (c) {
	    for (int i = 1; i <= 10; i++)
   		    if (i - 1 != c) d[i][0][0] = d[i - 1][0][0] + 1;
	        else d[i][0][0] = d[i - 1][0][0];

   		for (int i = c + 1; i <= 10; i++)
	        d[i][0][1] = 1;
	}
	else {
		for (int i = 1; i <= 9; i++)
			d[i][0][0] = 1;
		d[0][0][1] = 1;
	}

    for (int i = 1; i < 9; i++)
		for (int j = 0 + (c != 0); j <= 9; j++)
			for (int t = 0; t <= 9; t++)
				if (c)
					if (j == 1) {
						int x = 0;
						for (int L = i - 1; L >= 0; L--) {
							d[j][i][t] += d[9][L][t - x];

							if (c == 9) x++;
							if (x > t) break;
						}

						if (c != 9 && t == 0) d[j][i][t]++;
						if (c == 9 && t == i) d[j][i][t]++;
					}
					else {
						if (j == c + 1 && t == i + 1) d[j][i][t]++; //adaug numarul "ccccc" de i + 1 ori

						d[j][i][t] = d[j - 1][i][t];
						if (t - (j - 1 == c) >= 0) d[j][i][t] += d[1][i][t - (j - 1 == c)];
					}
				else 
					if (t - (j == 0) >= 0)
	                	for (int cif = 0; cif < 10; cif++)
							d[j][i][t] += d[cif][i - 1][t - (j == 0)];
}

int solve(int x) {
	if (x < 0) return 0;
	if (x == 0) return d[1][0][k];

	int cif[12] = {0,0,0,0,0,0,0,0,0,0,0}, sol = 0, nr = k;
	while (x) {
    	cif[++cif[0]] = x % 10;
		x /= 10;
	}

	if (c)
		for (int i = cif[0]; i > 0; i--) 
			if (i == 1) {
				if (cif[i] > 0)
					for (int j = nr; j <= 9; j++)
						sol = sol + d[cif[i] + 1][0][j];
			}
			else {
				for (int j = nr; j <= 9; j++)
	        		sol = sol + d[cif[i]][i - 1][j];
				if (cif[i] == c) nr--;
			}
	else 
		for (int i = cif[0]; i > 0; i--) {
            for (int j = 0 + (i == cif[0]); j < cif[i]; j++)
				for (int t = nr; t <= 9; t++)
					sol = sol + d[j][i - 1][t];
			if (cif[i] == 0) nr--;

			if (i < cif[0])
				for (int j = 1; j < 10; j++)
					for (int t = k; t <= 9; t++)
						sol = sol + d[j][i - 1][t];
		}

	if (!c && k <= 1) sol++;

	return sol;
}

int main() {
	freopen("cifre.in", "r", stdin);
	freopen("cifre.out", "w", stdout);

	scanf("%d %d %d %d", &a, &b, &c, &k);

	calc_din();

	printf("%.4lf\n", 1.0 * (solve(b) - solve(a - 1)) / (b - a + 1));

	return 0;
}