## Algorithm

B. Approximating a Constant Range
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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
Copy
`51 2 3 3 2`
output
Copy
`4`
input
Copy
`115 4 5 5 6 7 8 8 8 7 6`
output
Copy
`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].

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````#include <bits/stdc++.h>

using namespace std;

int const N = 1e5 + 1;
int n, a[N], fr[N];
queue<int> q;
set<int> st;

int main() {
scanf("%d", &n);
for(int i = 0; i < n; ++i)
scanf("%d", a + i);

int r = 0, res = 0, mn = 1e9, mx = -1e9;
while(r < n) {
++fr[a[r]];
q.push(a[r]);
st.insert(a[r]);

mn = min(mn, a[r]);
mx = max(mx, a[r]);

while(!q.empty() && fr[mx] && fr[mn] && mx - mn > 1) {
int tmp = q.front();
q.pop();
--fr[tmp];
}

if(fr[mn] == 0) {
st.erase(st.begin());
if(!st.empty());
mn = *st.begin();
}

if(fr[mx] == 0) {
st.erase(--st.end());
if(!st.empty())
mx = *(--st.end());
}

res = max(res, (int)q.size());
++r;
}

cout << res << endl;

return 0;
}``````
Copy The Code &

Input

cmd
5
1 2 3 3 2

Output

cmd
4