To help those contestants who struggle a lot in contests, the headquarters of Codeforces are planning to introduce Division 5. In this new division, the tags of all problems will be announced prior to the round to help the contestants.

The contest consists of n problems, where the tag of the i-th problem is denoted by an integer a_i.

You want to AK (solve all problems). To do that, you must solve the problems in some order. To make the contest funnier, you created extra limitations on yourself. You do not want to solve two problems consecutively with the same tag since it is boring. Also, you are afraid of big jumps in difficulties while solving them, so you want to minimize the number of times that you solve two problems consecutively that are not adjacent in the contest order.

Formally, your solve order can be described by a permutation p of length n. The cost of a permutation is defined as the number of indices i (1≤ i<n) where |p_{i+1}-p_i|>1. You have the requirement that a_{p_i}≠ a_{p_{i+1}} for all 1≤ i< n.

You want to know the minimum possible cost of permutation that satisfies the requirement. If no permutations meet this requirement, you should report about it.

Input

The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.

The first line of the description of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of problems in the contest.

The next line contains n integers a_1,a_2,… a_n (1 ≤ a_i ≤ n) — the tags of the problems.

It is guaranteed that the sum of n over all test cases does not exceed 10^5.

Output

For each test case, if there are no permutations that satisfy the required condition, print -1. Otherwise, print the minimum possible cost of a permutation that satisfies the required condition.

Example

Input


4
6
2 1 2 3 1 1
5
1 1 1 2 2
8
7 7 2 7 7 1 8 7
10
1 2 3 4 1 1 2 3 4 1


Output


1
3
-1
2

Note

In the first test case, let p=[5, 4, 3, 2, 1, 6]. The cost is 1 because we jump from p_5=1 to p_6=6, and |6-1|>1. This permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 1.

In the second test case, let p=[1,5,2,4,3]. The cost is 3 because |p_2-p_1|>1, |p_3-p_2|>1, and |p_4-p_3|>1. The permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 3.

In the third test case, for any order of solving the problems, we will solve two problems with the same tag consecutively, so the answer is -1.
#include <bits/stdc++.h>
using namespace std;
 
using ll = long long;
using ld = long double;
using db = double; 
using str = string; // yay python!

using pi = pair<int,int>;
using pl = pair<ll,ll>;
using pd = pair<db,db>;

using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>; 
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>; 
using vpd = vector<pd>;

#define tcT template<class T
#define tcTU tcT, class U
// ^ lol this makes everything look weird but I'll try it
tcT> using V = vector<T>; 
tcT, size_t SZ> using AR = array<T,SZ>; 
tcT> using PR = pair<T,T>;

// pairs
#define mp make_pair
#define f first
#define s second

// vectors
// oops size(x), rbegin(x), rend(x) need C++17
#define sz(x) int((x).size())
#define bg(x) begin(x)
#define all(x) bg(x), end(x)
#define rall(x) x.rbegin(), x.rend() 
#define sor(x) sort(all(x)) 
#define rsz resize
#define ins insert 
#define ft front()
#define bk back()
#define pb push_back
#define eb emplace_back 
#define pf push_front

#define lb lower_bound
#define ub upper_bound 
tcT> int lwb(V<T>& a, const T& b) { return int(lb(all(a),b)-bg(a)); }

// loops
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

const int MOD = 1e9+7; // 998244353;
const int MX = 2e5+5;
const ll INF = 1e18; // not too close to LLONG_MAX
const ld PI = acos((ld)-1);
const int dx[4] = {1,0,-1,0}, dy[4] = {0,1,0,-1}; // for every grid problem!!
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count()); 
template<class T> using pqg = priority_queue<T,vector<T>,greater<T>>;

// bitwise ops
// also see https://gcc.gnu.org/onlinedocs/gcc/Other-Builtins.html
constexpr int pct(int x) { return __builtin_popcount(x); } // # of bits set
constexpr int bits(int x) { assert(x >= 0);
	return x == 0 ? 0 : 31-__builtin_clz(x); } // floor(log2(x)) 
constexpr int p2(int x) { return 1<<x; }
constexpr int msk2(int x) { return p2(x)-1; }

ll cdiv(ll a, ll b) { return a/b+((a^b)>0&&a%b); } // divide a by b rounded up
ll fdiv(ll a, ll b) { return a/b-((a^b)<0&&a%b); } // divide a by b rounded down

tcT> bool ckmin(T& a, const T& b) {
	return b < a ? a = b, 1 : 0; } // set a = min(a,b)
tcT> bool ckmax(T& a, const T& b) {
	return a < b ? a = b, 1 : 0; }

tcTU> T fstTrue(T lo, T hi, U f) {
	hi ++; assert(lo <= hi); // assuming f is increasing
	while (lo < hi) { // find first index such that f is true 
		T mid = lo+(hi-lo)/2;
		f(mid) ? hi = mid : lo = mid+1; 
	} 
	return lo;
}
tcTU> T lstTrue(T lo, T hi, U f) {
	lo --; assert(lo <= hi); // assuming f is decreasing
	while (lo < hi) { // find first index such that f is true 
		T mid = lo+(hi-lo+1)/2;
		f(mid) ? lo = mid : hi = mid-1;
	} 
	return lo;
}
tcT> void remDup(vector<T>& v) { // sort and remove duplicates
	sort(all(v)); v.erase(unique(all(v)),end(v)); }
tcTU> void erase(T& t, const U& u) { // don't erase
	auto it = t.find(u); assert(it != end(t));
	t.erase(it); } // element that doesn't exist from (multi)set

// INPUT
#define tcTUU tcT, class ...U
tcT> void re(complex<T>& c);
tcTU> void re(pair<T,U>& p);
tcT> void re(V<T>& v);
tcT, size_t SZ> void re(AR<T,SZ>& a);

tcT> void re(T& x) { cin >> x; }
void re(db& d) { str t; re(t); d = stod(t); }
void re(ld& d) { str t; re(t); d = stold(t); }
tcTUU> void re(T& t, U&... u) { re(t); re(u...); }

tcT> void re(complex<T>& c) { T a,b; re(a,b); c = {a,b}; }
tcTU> void re(pair<T,U>& p) { re(p.f,p.s); }
tcT> void re(V<T>& x) { trav(a,x) re(a); }
tcT, size_t SZ> void re(AR<T,SZ>& x) { trav(a,x) re(a); }
tcT> void rv(int n, V<T>& x) { x.rsz(n); re(x); }

// TO_STRING
#define ts to_string
str ts(char c) { return str(1,c); }
str ts(const char* s) { return (str)s; }
str ts(str s) { return s; }
str ts(bool b) { 
	#ifdef LOCAL
		return b ? "true" : "false"; 
	#else 
		return ts((int)b);
	#endif
}
tcT> str ts(complex<T> c) { 
	stringstream ss; ss << c; return ss.str(); }
str ts(V<bool> v) {
	str res = "{"; F0R(i,sz(v)) res += char('0'+v[i]);
	res += "}"; return res; }
template<size_t SZ> str ts(bitset<SZ> b) {
	str res = ""; F0R(i,SZ) res += char('0'+b[i]);
	return res; }
tcTU> str ts(pair<T,U> p);
tcT> str ts(T v) { // containers with begin(), end()
	#ifdef LOCAL
		bool fst = 1; str res = "{";
		for (const auto& x: v) {
			if (!fst) res += ", ";
			fst = 0; res += ts(x);
		}
		res += "}"; return res;
	#else
		bool fst = 1; str res = "";
		for (const auto& x: v) {
			if (!fst) res += " ";
			fst = 0; res += ts(x);
		}
		return res;

	#endif
}
tcTU> str ts(pair<T,U> p) {
	#ifdef LOCAL
		return "("+ts(p.f)+", "+ts(p.s)+")"; 
	#else
		return ts(p.f)+" "+ts(p.s);
	#endif
}

// OUTPUT
tcT> void pr(T x) { cout << ts(x); }
tcTUU> void pr(const T& t, const U&... u) { 
	pr(t); pr(u...); }
void ps() { pr("\n"); } // print w/ spaces
tcTUU> void ps(const T& t, const U&... u) { 
	pr(t); if (sizeof...(u)) pr(" "); ps(u...); }

// DEBUG
void DBG() { cerr << "]" << endl; }
tcTUU> void DBG(const T& t, const U&... u) {
	cerr << ts(t); if (sizeof...(u)) cerr << ", ";
	DBG(u...); }
#ifdef LOCAL // compile with -DLOCAL, chk -> fake assert
	#define dbg(...) cerr << "Line(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
	#define chk(...) if (!(__VA_ARGS__)) cerr << "Line(" << __LINE__ << ") -> function(" \
		 << __FUNCTION__  << ") -> CHK FAILED: (" << #__VA_ARGS__ << ")" << "\n", exit(0);
#else
	#define dbg(...) 0
	#define chk(...) 0
#endif

void setPrec() { cout << fixed << setprecision(15); }
void unsyncIO() { cin.tie(0)->sync_with_stdio(0); }
// FILE I/O
void setIn(str s) { freopen(s.c_str(),"r",stdin); }
void setOut(str s) { freopen(s.c_str(),"w",stdout); }
void setIO(str s = "") {
	unsyncIO(); setPrec();
	// cin.exceptions(cin.failbit); 
	// throws exception when do smth illegal
	// ex. try to read letter into int
	if (sz(s)) setIn(s+".in"), setOut(s+".out"); // for USACO
}

// make sure to intialize ALL GLOBAL VARS between tcs!

int N;

void solve(int tc) {
	re(N);
	vi A(N); re(A);
	vpi inter;
	for (int i = 0; i < N; ) {
		int st = i;
		while (i+1 < N && A[i+1] != A[i]) ++i;
		inter.pb({st,i});
		// dbg(st,i);
		++ i;
	}
	map<int,int> cnt;
	trav(t,inter) {
		++ cnt[A[t.f]];
		++ cnt[A[t.s]];
	}
	dbg(cnt);
	vpi oc;
	trav(t,cnt) oc.pb({t.s,t.f});
	sort(rall(oc));
	dbg(oc);
	int tot = 2*sz(inter);
	int ans = sz(inter)-1;
	if (oc[0].f <= tot-oc[0].f+2) {
		ps(ans);
		return;
	}
	trav(q,inter) {
		FOR(i,q.f,q.s) if (A[i] != oc[0].s && A[i+1] != oc[0].s) {
			ans ++; tot += 2;
			if (oc[0].f <= tot-oc[0].f+2) {
				ps(ans);
				return;
			}
		}
	}
	ps(-1);
	// ps("OOPS");

}

int main() {
	setIO();
	int TC; re(TC);
	FOR(i,1,TC+1) solve(i);
}

/* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
	* DON'T GET STUCK ON ONE APPROACH
*/
