When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?

You're given a sequence of n data points a1, ..., an. There aren't any big jumps between consecutive data points — for each 1 ≤ i < n, it's guaranteed that |ai + 1 - ai| ≤ 1.

A range [l, r] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let M be the maximum and m the minimum value of ai for l ≤ i ≤ r; the range [l, r] is almost constant if M - m ≤ 1.

Find the length of the longest almost constant range.

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of data points.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000).

Output

Print a single number — the maximum length of an almost constant range of the given sequence.

Examples

Input

5
1 2 3 3 2


Output

4


Input

11
5 4 5 5 6 7 8 8 8 7 6


Output

5

Note

In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.

In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10].


import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Optional;
import java.util.Scanner;

public class ApproximatingAConstantRange {

	private static class Range {

		private final int start;

		private final int offset;

		private Range(int start, int offset) {
			this.start = start;
			this.offset = offset;
		}

		public Optional<Range> include(int n) {
			if (n == start || n == start + offset) {
				return Optional.of(this);
			}
			if (offset == 0) {
				if (n == start + 1) {
					return Optional.of(new Range(start, 1));
				} else if (n == start - 1) {
					return Optional.of(new Range(start, -1));
				}
			}
			return Optional.empty();
		}

		@Override
		public int hashCode() {
			final int prime = 31;
			int result = 1;
			result = prime * result + offset;
			result = prime * result + start;
			return result;
		}

		@Override
		public boolean equals(Object obj) {
			if (this == obj)
				return true;
			if (obj == null)
				return false;
			if (getClass() != obj.getClass())
				return false;
			Range other = (Range) obj;
			if (offset != other.offset)
				return false;
			if (start != other.start)
				return false;
			return true;
		}

		public static Range start(int start) {
			return new Range(start, 0);
		}

	}

	public static void main(String[] args) {
		try (Scanner s = new Scanner(System.in)) {
			int n = s.nextInt();
			// Store the current active ranges and their durations.
			Map<Range, Integer> currentRangeDurations = new HashMap<>();
			int bestRange = 1;
			for (int i = 0; i < n; i++) {
				int current = s.nextInt();
				// Populate a new map to avoid concurrent modification.
				Map<Range, Integer> tempStorage = new HashMap<>();
				for (Entry<Range, Integer> entry : currentRangeDurations.entrySet()) {
					int duration = entry.getValue();
					Optional<Range> included = entry.getKey().include(current);
					// Check if we can add our current index to this range.
					if (included.isPresent()) {
						duration++;
						Integer existingDuration = tempStorage.get(included.get());
						if (existingDuration != null) {
							tempStorage.put(included.get(), Math.max(existingDuration, duration));
						} else {
							tempStorage.put(included.get(), duration);
						}
					}
					// Update our best range duration we've seen so far.
					bestRange = Math.max(bestRange, duration);
				}
				Range currentRange = Range.start(current);
				if (!currentRangeDurations.containsKey(currentRange)) {
					tempStorage.put(currentRange, 1);
				}
				currentRangeDurations = tempStorage;
			}
			System.out.println(bestRange);
		}
	}

}
