British mathematician John Littlewood once said about Indian mathematician Srinivasa Ramanujan that "every positive integer was one of his personal friends."

It turns out that positive integers can also be friends with each other! You are given an array a of distinct positive integers. 

Define a subarray a_i, a_{i+1}, …, a_j to be a friend group if and only if there exists an integer m ≥ 2 such that a_i mod m = a_{i+1} mod m = … = a_j mod m, where x mod y denotes the remainder when x is divided by y.

Your friend Gregor wants to know the size of the largest friend group in a.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 2⋅ 10^4). 

Each test case begins with a line containing the integer n (1 ≤ n ≤ 2 ⋅ 10^5), the size of the array a.

The next line contains n positive integers a_1, a_2, …, a_n (1 ≤ a_i ≤ {10}^{18}), representing the contents of the array a. It is guaranteed that all the numbers in a are distinct.

It is guaranteed that the sum of n over all test cases is less than 2⋅ 10^5.

Output

Your output should consist of t lines. Each line should consist of a single integer, the size of the largest friend group in a.

Example

Input


4
5
1 5 2 4 6
4
8 2 5 10
2
1000 2000
8
465 55 3 54 234 12 45 78


Output


3
3
2
6

Note

In the first test case, the array is [1,5,2,4,6]. The largest friend group is [2,4,6], since all those numbers are congruent to 0 modulo 2, so m=2.

In the second test case, the array is [8,2,5,10]. The largest friend group is [8,2,5], since all those numbers are congruent to 2 modulo 3, so m=3.

In the third case, the largest friend group is [1000,2000]. There are clearly many possible values of m that work.
1
3
1 137438953474 274877906947
 5
1
13
1
1
6
7 19 28 25 4 10
3
3 19 10
9
26 1 28 8 24 27 12 16 17
 1
69
2 4 10 16 22 28 34 40 46 52 58 64 70 76 82 88 94 100 106 112 118 124 130 136 142 148 154 160 166 172 178 184 190 196 202 208 214 220 226 232 238 244 250 256 262 268 274 280 286 292 298 304 310 316 322 328 334 340 346 352 358 364 370 376 382 388 391 397 403
 1
20
1 3 5 7 9 11 13 15 17 19 22 25 28 31 34 37 40 43 46 49
 1
20
16 15 17 8 30 23 20 28 27 6 1 18 24 2 10 5 14 29 12 7
 3
3
14 4 27
13
26 14 6 10 5 3 29 2 24 12 22 11 1
4
10 25 29 11
 1
4
1 140688977775057788 281377955550115575 281377955925200962

3
 1
1
6
2
3
 68
 11
 4
 2
4
3
 4

//BY: YASH JAIN, CF: BitSane
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> // Common file 
#include <ext/pb_ds/tree_policy.hpp>
#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
using namespace __gnu_pbds;
using namespace std;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
#define May_the_fork_be_with_you ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define all(x) (x).begin(), (x).end()
#define rall(v) v.rbegin(),v.rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define deb(x) cout<<#x<<": "<<x<<endl;
#define debn(x) cout<<#x<<": "<<x<<"  ";
#define MOD 1000000007
#define mod 998244353
#define INF 1e18
#define ll long long
#define f first
#define s second
#define pb push_back
#define eb emplace_back
#define endl "\n"
#define int long long
#define sq(x) ((x)*(x))
typedef vector<int> vi;
ll gcd(ll a, ll b) {if (!b)return a; return gcd(b, a % b);}
ll power(ll x, ll y, ll p = INF) {ll res = 1; x %= p; while (y > 0) {if (y & 1)res = (res * x) % p; y = y >> 1; x = (x * x) % p;} return res;}
// Do Not use power when calculating powers of 2 (it's inefficient)

// This is lain's segtree template
template<class T>
struct SegTree {
    int n;
    vector<T> t;
    T E = 0;

    T comb(T a, T b) { // Segtree function
        return gcd(a, b);
    }
    SegTree() {}
    SegTree(int _n) : n(_n), t(4 * n + 5) {}
    SegTree(vector<T>& bld) : n(bld.size()), t(4 * n + 5) {
        build(bld, 1, 0, n - 1);
    }
    void build(vector<T>& bld, int node, int tl, int tr) {
        if (tl == tr) {
            t[node] = bld[tl];
        }
        else {
            int tm = (tl + tr) / 2;
            build(bld, 2 * node, tl, tm);
            build(bld, 2 * node + 1, tm + 1, tr);
            pull(node);
        }
    }
    void pull(int node) { t[node] = comb(t[2 * node], t[2 * node + 1]); }
    T query(int l, int r) {return query(l, r, 1, 0, n - 1);}
    T query(int l, int r, int node, int tl, int tr) {
        if (r < tl || tr < l) return E;
        if (l <= tl && tr <= r) return t[node];
        int tm = (tl + tr) / 2;
        return comb(query(l, r, 2 * node, tl, tm), query(l, r, 2 * node + 1, tm + 1, tr));
    }
    void update(int p, T val) { return update(p, val, 1, 0, n - 1); }
    void update(int p, T val, int node, int tl, int tr) {
        if (tl == tr) {
            t[node] = val;
            return;
        }
        int tm = (tl + tr) / 2;
        if (p <= tm) update(p, val, 2 * node, tl, tm);
        else update(p, val, 2 * node + 1, tm + 1, tr);
        pull(node);
    }
};

void solve() {
    int n; cin >> n;
    if (n == 1) {
        cout << 1 << endl;
        return;
    }
    vi a(n), b(n - 1);
    forn(i, n) cin >> a[i];
    forn(i, n - 1) b[i] = abs(a[i] - a[i + 1]);
    SegTree<int> st(b);
    int i = 0, j = 0, res = 0;
    while (j < n - 1) {
        if (st.query(i, j) != 1) {
            res = max(res, j - i + 2);
            j++;
        }
        else i++;
    }
    cout << res << endl;
}

int32_t main() {

#ifndef ONLINE_JUDGE
    // for geting input form input.txt
    freopen("input.txt", "r", stdin);
    // for wrting output to output.txt
    freopen("output.txt", "w", stdout);
#endif

    May_the_fork_be_with_you
    int t = 1;
    int x = 1;
    cin >> t;
    cout << fixed << setprecision(12);
    while (t--) {
        // cout << "Case #" << x << ": ";
        solve();
        x++;
    }

#ifndef ONLINE_JUDGE
    cerr << "Time Taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
#endif
}