Recently, Tokitsukaze found an interesting game. Tokitsukaze had n items at the beginning of this game. However, she thought there were too many items, so now she wants to discard m (1 ≤ m ≤ n) special items of them.

These n items are marked with indices from 1 to n. In the beginning, the item with index i is placed on the i-th position. Items are divided into several pages orderly, such that each page contains exactly k positions and the last positions on the last page may be left empty.

Tokitsukaze would do the following operation: focus on the first special page that contains at least one special item, and at one time, Tokitsukaze would discard all special items on this page. After an item is discarded or moved, its old position would be empty, and then the item below it, if exists, would move up to this empty position. The movement may bring many items forward and even into previous pages, so Tokitsukaze would keep waiting until all the items stop moving, and then do the operation (i.e. check the special page and discard the special items) repeatedly until there is no item need to be discarded.

<image> Consider the first example from the statement: n=10, m=4, k=5, p=[3, 5, 7, 10]. The are two pages. Initially, the first page is special (since it is the first page containing a special item). So Tokitsukaze discards the special items with indices 3 and 5. After, the first page remains to be special. It contains [1, 2, 4, 6, 7], Tokitsukaze discards the special item with index 7. After, the second page is special (since it is the first page containing a special item). It contains [9, 10], Tokitsukaze discards the special item with index 10.

Tokitsukaze wants to know the number of operations she would do in total.

Input

The first line contains three integers n, m and k (1 ≤ n ≤ 10^{18}, 1 ≤ m ≤ 10^5, 1 ≤ m, k ≤ n) — the number of items, the number of special items to be discarded and the number of positions in each page.

The second line contains m distinct integers p_1, p_2, …, p_m (1 ≤ p_1 < p_2 < … < p_m ≤ n) — the indices of special items which should be discarded.

Output

Print a single integer — the number of operations that Tokitsukaze would do in total.

Examples

Input


10 4 5
3 5 7 10


Output


3


Input


13 4 5
7 8 9 10


Output


1

Note

For the first example:

  * In the first operation, Tokitsukaze would focus on the first page [1, 2, 3, 4, 5] and discard items with indices 3 and 5; 
  * In the second operation, Tokitsukaze would focus on the first page [1, 2, 4, 6, 7] and discard item with index 7; 
  * In the third operation, Tokitsukaze would focus on the second page [9, 10] and discard item with index 10. 



For the second example, Tokitsukaze would focus on the second page [6, 7, 8, 9, 10] and discard all special items at once.
import java.util.*;
import java.io.*;
import java.math.*;

/*
10000000 99999 1

10 10 1
1 2 3 4 5 6 7 8 9 10
 */

public class round573d2c {

	public static void main(String args[]) {
		FastScanner in = new FastScanner(System.in);

		long n = in.nextLong();
		int m = in.nextInt();
		long k = in.nextLong();
		long[] p = new long[m];
		String line = in.nextLine();
		StringTokenizer st = new StringTokenizer(line);
		//System.out.println("here");
		for (int i = 0; i < m ;i++) {
			//p[i] = in.nextLong();
			p[i] = Long.parseLong(st.nextToken());
		}
		//System.out.println("here2");
		int i = 0;
		long cap = 1;
		int stored = 0;
		int ans = 0;
		while (i < m) {
			//System.out.println(cap);
			if (p[i] >= cap) {
				if (stored != 0) {
					ans++;
					cap += stored;
					stored = 0;
				}
				else {
					stored = 0;
					cap += ((p[i] - cap)/k + 1)*k;
					//cap += k;
				}
				
			}
			else {
				i++;
				stored++;
			}
		}
		if (stored != 0) {
			ans++;
		}
		System.out.println(ans);
		//in.close();
	}

	// ======================================================================================
	// =============================== Reference Code =======================================
	// ======================================================================================

	static int greatestDivisor(int n) {
		int limit = (int) Math.sqrt(n);
		int max = 1;
		for (int i = 2; i <= limit; i++) {
			if (n % i == 0) {
				max = Integer.max(max, i);
				max = Integer.max(max, n / i);
			}
		}
		return max;
	}

	// Method to return all primes smaller than or equal to 
	// n using Sieve of Eratosthenes 
	static boolean[] sieveOfEratosthenes(int n) {
		// Create a boolean array "prime[0..n]" and initialize 
		// all entries it as true. A value in prime[i] will 
		// finally be false if i is Not a prime, else true. 
		boolean prime[] = new boolean[n + 1];
		for (int i = 0; i <= n; i++)
			prime[i] = true;

		prime[0] = false;
		prime[1] = false;

		for (int p = 2; p * p <= n; p++) {
			// If prime[p] is not changed, then it is a prime 
			if (prime[p] == true) {
				// Update all multiples of p 
				for (int i = p * p; i <= n; i += p)
					prime[i] = false;
			}
		}

		return prime;
	}

	// Binary search for number greater than or equal to target
	// returns -1 if number not found
	private static int bin_gteq(int[] a, int key) {
		int low = 0;
		int high = a.length;
		int max_limit = high;
		while (low < high) {
			int mid = low + (high - low) / 2;
			if (a[mid] < key) {
				low = mid + 1;
			} else
				high = mid;
		}

		return high == max_limit ? -1 : high;
	}

	public static int gcd(int a, int b) {
		if (a == 0)
			return b;

		return gcd(b % a, a);
	}

	static class Tuple<X, Y> {
		public final X x;
		public final Y y;

		public Tuple(X x, Y y) {
			this.x = x;
			this.y = y;
		}

		public String toString() {
			return "(" + x + "," + y + ")";
		}
	}

	static class Tuple3<X, Y, Z> {
		public final X x;
		public final Y y;
		public final Z z;

		public Tuple3(X x, Y y, Z z) {
			this.x = x;
			this.y = y;
			this.z = z;
		}

		public String toString() {
			return "(" + x + "," + y + "," + z + ")";
		}
	}

	static Tuple3<Integer, Integer, Integer> gcdExtended(int a, int b, int x, int y) {
		// Base Case 
		if (a == 0) {
			x = 0;
			y = 1;
			return new Tuple3(0, 1, b);
		}

		int x1 = 1, y1 = 1; // To store results of recursive call 
		Tuple3<Integer, Integer, Integer> tuple = gcdExtended(b % a, a, x1, y1);
		int gcd = tuple.z;
		x1 = tuple.x;
		y1 = tuple.y;

		// Update x and y using results of recursive 
		// call 
		x = y1 - (b / a) * x1;
		y = x1;

		return new Tuple3(x, y, gcd);
	}

	// Returns modulo inverse of a  
	// with respect to m using extended 
	// Euclid Algorithm. Refer below post for details: 
	// https://www.geeksforgeeks.org/multiplicative-inverse-under-modulo-m/ 
	static int inv(int a, int m) {
		int m0 = m, t, q;
		int x0 = 0, x1 = 1;

		if (m == 1)
			return 0;

		// Apply extended Euclid Algorithm 
		while (a > 1) {
			// q is quotient 
			q = a / m;

			t = m;

			// m is remainder now, process 
			// same as euclid's algo 
			m = a % m;
			a = t;

			t = x0;

			x0 = x1 - q * x0;

			x1 = t;
		}

		// Make x1 positive 
		if (x1 < 0)
			x1 += m0;

		return x1;
	}

	// k is size of num[] and rem[]. 
	// Returns the smallest number 
	// x such that: 
	// x % num[0] = rem[0], 
	// x % num[1] = rem[1], 
	// .................. 
	// x % num[k-2] = rem[k-1] 
	// Assumption: Numbers in num[] are pairwise  
	// coprime (gcd for every pair is 1) 
	static int findMinX(int num[], int rem[], int k) {
		// Compute product of all numbers 
		int prod = 1;
		for (int i = 0; i < k; i++)
			prod *= num[i];

		// Initialize result 
		int result = 0;

		// Apply above formula 
		for (int i = 0; i < k; i++) {
			int pp = prod / num[i];
			result += rem[i] * inv(pp, num[i]) * pp;
		}

		return result % prod;
	}

	/**
	 * Source: Matt Fontaine
	 */
	static class FastScanner {
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int chars;

		public FastScanner(InputStream stream) {
			this.stream = stream;
		}

		int read() {
			if (chars == -1)
				throw new InputMismatchException();
			if (curChar >= chars) {
				curChar = 0;
				try {
					chars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (chars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		boolean isEndline(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}

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

		public long nextLong() {
			return Long.parseLong(next());
		}

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

		public String next() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public String nextLine() {
			int c = read();
			while (isEndline(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndline(c));
			return res.toString();
		}
	}

}