Problem: There used to be unrest in the Galactic Senate. Several thousand solar systems had declared their intentions to leave the Republic. But fear not! Master Heidi was able to successfully select the Jedi Knights that have restored peace in the galaxy. However, she knows that evil never sleeps and a time may come when she will need to pick another group of Jedi Knights. She wants to be sure she has enough options to do so.

There are n Jedi Knights, each of them with a lightsaber of one of m colors. Given a number k, compute the number of differently colored collections of k lightsabers that some k Jedi Knights might have. Jedi Knights with lightsabers of the same color are indistinguishable (it's not the person, it's the lightsaber color that matters!), and their order does not matter; that is, we consider two collections of Jedi Knights to be different if and only if their vectors of counts of lightsabers of each color (like what you were given in the easy and the medium versions) are different. We count all subsets, not only contiguous subsegments of the input sequence. Output the answer modulo 1009.

Input

The first line of the input contains n (1 ≤ n ≤ 2·105), m (1 ≤ m ≤ n) and k (1 ≤ k ≤ n). The second line contains n integers in the range {1, 2, ..., m} representing colors of the lightsabers of subsequent Jedi Knights.

Output

Output one number: the number of differently colored collections of k lightsabers modulo 1009.

Example

Input

4 3 2
1 2 3 2


Output

4

Solution (in JAVA):

Solution:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.text.DecimalFormat;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.PriorityQueue;
import java.util.Random;
import java.util.StringTokenizer;

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
/**
 *
 * @author Andy Phan
 */
public class f {
    static long MOD = 1009;

    public static void main(String[] args) {
        new f().solve();
    }

    public void solve() {
        FS in = new FS(System.in);
        PrintWriter out = new PrintWriter(System.out);
        
        int n = in.nextInt();
        int m = in.nextInt();
        int k = in.nextInt();
        int[] num = new int[m];
        for(int i = 0; i < n; i++) num[in.nextInt()-1]++;
        PriorityQueue<long[]> q = new PriorityQueue<long[]>((a, b) -> a.length-b.length);
        for(int i = 0; i < m; i++) {
            long[] tmp = new long[num[i]+1];
            Arrays.fill(tmp, 1);
            q.add(tmp);
        }
        while(q.size() > 1) {
            q.add(FFT.multiply(q.poll(), q.poll()));
        }
        System.out.println(q.poll()[k]);
        out.close();
    }

    // SET maxk appropriately!!! ~log(maxn) //%
    static class FFT {

        static final int maxk = 19, maxn = (1 << maxk) + 1;
        // Init: wR, wI, rR, rI, aR, aI   to   new double[maxn] !!!
        //#
        static double[] wR = new double[maxn],
            wI = new double[maxn],
            rR = new double[maxn],
            rI = new double[maxn],
            aR = new double[maxn],
            aI = new double[maxn]; //$
        static int n, k, lastk = -1, dp[] = new int[maxn];

        static void fft(boolean inv) {
            if (lastk != k) {
                lastk = k;
                dp[0] = 0;
                for (int i = 1, g = -1; i < n; i++) {
                    if ((i & (i - 1)) == 0) {
                        g++;
                    }
                    dp[i] = dp[i ^ (1 << g)] ^ (1 << (k - 1 - g));
                }
                wR[1] = 1;
                wI[1] = 0;
                for (int t = 0; t < k - 1; t++) {
                    double a = Math.PI / n * (1 << (k - 1 - t));
                    double curR = Math.cos(a), curI = Math.sin(a);
                    int p2 = (1 << t), p3 = p2 * 2;
                    for (int j = p2, k = j * 2; j < p3; j++, k += 2) {
                        wR[k] = wR[j];
                        wI[k] = wI[j];
                        wR[k + 1] = wR[j] * curR - wI[j] * curI;
                        wI[k + 1] = wR[j] * curI + wI[j] * curR;
                    }
                }
            }
            for (int i = 0; i < n; i++) {
                int d = dp[i];
                if (i < d) {
                    double tmp = aR[i];
                    aR[i] = aR[d];
                    aR[d] = tmp;
                    tmp = aI[i];
                    aI[i] = aI[d];
                    aI[d] = tmp;
                }
            }
            if (inv) {
                for (int i = 0; i < n; i++) {
                    aI[i] = -aI[i];
                }
            }
            for (int len = 1; len < n; len <<= 1) {
                for (int i = 0; i < n; i += len) {
                    int wit = len;
                    for (int it = 0, j = i + len; it < len; it++, i++, j++) {
                        double tmpR = aR[j] * wR[wit] - aI[j] * wI[wit];
                        double tmpI = aR[j] * wI[wit] + aI[j] * wR[wit];
                        wit++;
                        aR[j] = aR[i] - tmpR;
                        aI[j] = aI[i] - tmpI;
                        aR[i] += tmpR;
                        aI[i] += tmpI;
                    }
                }
            }
        }

        static long[] multiply(long[] a, long[] b) {
            int na = a.length, nb = b.length;
            for (k = 0, n = 1; n < na + nb - 1; n <<= 1, k++) {
            }
            for (int i = 0; i < n; ++i) {
                aR[i] = i < na ? a[i] : 0;
                aI[i] = i < nb ? b[i] : 0;
            }
            fft(false);
            aR[n] = aR[0];
            aI[n] = aI[0];
            double q = -1.0 / n / 4.0;
            for (int i = 0; i <= n - i; ++i) {
                double tmpR = aR[i] * aR[i] - aI[i] * aI[i];
                double tmpI = aR[i] * aI[i] * 2;
                tmpR -= aR[n - i] * aR[n - i] - aI[n - i] * aI[n - i];
                tmpI -= aR[n - i] * aI[n - i] * -2;
                aR[i] = -tmpI * q;
                aI[i] = tmpR * q;
                aR[n - i] = aR[i];
                aI[n - i] = -aI[i];
            }
            fft(true);
            long[] ans = new long[n = na + nb - 1]; // ONLY MOD IF NEEDED
            for (int i = 0; i < n; i++) {
                ans[i] = Math.round(aR[i]) % MOD;
            }
            return ans;
        }

        static void fft2(double[][] xr, double[][] xi, boolean inv) {
            n = xr[0].length;
            k = Integer.numberOfTrailingZeros(n);
            for (int i = 0; i < xr.length; i++) {
                for (int j = 0; j < n; j++) {
                    aR[j] = xr[i][j];
                    aI[j] = xi[i][j];
                }
                fft(inv);
                for (int j = 0; j < n; j++) {
                    xr[i][j] = aR[j] / (inv ? n : 1);
                    xi[i][j] = aI[j] / (inv ? -n : 1);
                }
            }
            n = xr.length;
            k = Integer.numberOfTrailingZeros(n);
            for (int j = 0; j < xr[0].length; j++) {
                for (int i = 0; i < n; i++) {
                    aR[i] = xr[i][j];
                    aI[i] = xi[i][j];
                }
                fft(inv);
                for (int i = 0; i < n; i++) {
                    xr[i][j] = aR[i] / (inv ? n : 1);
                    xi[i][j] = aI[i] / (inv ? -n : 1);
                }
            }
        }

        static long[][] multiply2(long[][] a, long[][] b) {
            int n1, n2;
            for (n1 = 1; n1 < a.length + b.length - 1; n1 <<= 1) {
            }
            for (n2 = 1; n2 < a[0].length + b[0].length - 1; n2 <<= 1) {
            }
            double[][] ar = new double[n1][n2], ai = new double[n1][n2];
            double[][] br = new double[n1][n2], bi = new double[n1][n2];
            for (int i = 0; i < a.length; i++) {
                for (int j = 0; j < a[i].length; j++) {
                    ar[i][j] = a[i][j];
                }
            }
            for (int i = 0; i < b.length; i++) {
                for (int j = 0; j < b[i].length; j++) {
                    br[i][j] = b[i][j];
                }
            }
            fft2(ar, ai, false);
            fft2(br, bi, false);
            for (int i = 0; i < n1; i++) {
                for (int j = 0; j < n2; j++) {
                    double r1 = ar[i][j], r2 = br[i][j];
                    double i1 = ai[i][j], i2 = bi[i][j];
                    double real = r1 * r2 - i1 * i2;
                    ai[i][j] = i1 * r2 + i2 * r1;
                    ar[i][j] = real;
                }
            }
            fft2(ar, ai, true);
            long[][] result = new long[n1 = a.length + b.length - 1][n2 = a[0].length + b[0].length - 1];
            for (int i = 0; i < n1; i++) {
                for (int j = 0; j < n2; j++) {
                    result[i][j] = Math.round(ar[i][j]);
                }
            }
            return result;
        }
        //#

        static long[] multiplyPrecise(long[] a, long[] b) {
            long k = (long) (Math.sqrt(MOD));
            long[] a1 = new long[a.length], a2 = new long[a.length];
            long[] b1 = new long[b.length], b2 = new long[b.length];
            for (int i = 0; i < a.length; i++) {
                a1[i] = a[i] % k;
                a2[i] = a[i] / k;
            }
            for (int i = 0; i < b.length; i++) {
                b1[i] = b[i] % k;
                b2[i] = b[i] / k;
            }
            long[] r11 = multiply(a1, b1), r12 = multiply(a1, b2), r21 = multiply(a2, b1), r22 = multiply(a2, b2);
            long[] res = new long[r11.length];
            for (int i = 0; i < res.length; i++) {
                res[i] = (k * k * r22[i] + k * (r12[i] + r21[i]) + r11[i]) % MOD;
            }
            return res;
        }
        //$
        //#

        static long[] multiplyOr(long[] eq1, long[] eq2) {
            int n = Math.max(eq1.length, eq2.length);
            if ((n & (n - 1)) != 0) {
                n = Integer.highestOneBit(n) * 2;
            }
            eq1 = fftOr(eq1, n, false);
            eq2 = fftOr(eq2, n, false);
            for (int i = 0; i < eq1.length; i++) {
                eq1[i] *= eq2[i];
            }
            eq1 = fftOr(eq1, n, true);
            return eq1;
        }
        //$
        // To use: FFT both, product, iFFT (n is next power of 2)

        static long[] fftOr(long[] arr, int n, boolean invert) {
            long[] ans = Arrays.copyOf(arr, n);
            for (int b = 1; b < n; b <<= 1) {
                for (int i = 0; i < n; i++) {
                    if ((i & b) != 0) {
                        continue;
                    }
                    ans[i + b] += invert ? -ans[i] : ans[i];
                }
            }
            return ans;
        }

        static long[] fftXor(long[] arr, int n, boolean invert) {
            long[] ans = Arrays.copyOf(arr, n);
            for (int b = 1; b < n; b <<= 1) {
                for (int i = 0; i < n; i++) {
                    if ((i & b) != 0) {
                        continue;
                    }
                    long u = ans[i], v = ans[i + b];
                    ans[i] = u + v;
                    ans[i + b] = u - v;
                }
            }
            if (invert) {
                for (int i = 0; i < n; i++) {
                    ans[i] /= n;
                }
            }
            return ans;
        }
    }

    static class FS {

        BufferedReader in;
        StringTokenizer token;

        public FS(InputStream str) {
            in = new BufferedReader(new InputStreamReader(str));
        }

        public String next() {
            if (token == null || !token.hasMoreElements()) {
                try {
                    token = new StringTokenizer(in.readLine());
                } catch (IOException ex) {
                }
                return next();
            }
            return token.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}