You are given an array a[0 … n - 1] = [a_0, a_1, …, a_{n - 1}] of zeroes and ones only. Note that in this problem, unlike the others, the array indexes are numbered from zero, not from one.

In one step, the array a is replaced by another array of length n according to the following rules: 

  1. First, a new array a^{→ d} is defined as a cyclic shift of the array a to the right by d cells. The elements of this array can be defined as a^{→ d}_i = a_{(i + n - d) mod n}, where (i + n - d) mod n is the remainder of integer division of i + n - d by n. 

It means that the whole array a^{→ d} can be represented as a sequence $$$a^{→ d} = [a_{n - d}, a_{n - d + 1}, …, a_{n - 1}, a_0, a_1, …, a_{n - d - 1}]$$$

  2. Then each element of the array a_i is replaced by a_i  \&  a^{→ d}_i, where \& is a logical "AND" operator. 



For example, if a = [0, 0, 1, 1] and d = 1, then a^{→ d} = [1, 0, 0, 1] and the value of a after the first step will be [0  \&  1, 0  \&  0, 1  \&  0, 1  \&  1], that is [0, 0, 0, 1].

The process ends when the array stops changing. For a given array a, determine whether it will consist of only zeros at the end of the process. If yes, also find the number of steps the process will take before it finishes.

Input

The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases.

The next 2t lines contain descriptions of the test cases. 

The first line of each test case description contains two integers: n (1 ≤ n ≤ 10^6) — array size and d (1 ≤ d ≤ n) — cyclic shift offset. The second line of the description contains n space-separated integers a_i (0 ≤ a_i ≤ 1) — elements of the array.

It is guaranteed that the sum of n over all test cases does not exceed 10^6.

Output

Print t lines, each line containing the answer to the corresponding test case. The answer to a test case should be a single integer — the number of steps after which the array will contain only zeros for the first time. If there are still elements equal to 1 in the array after the end of the process, print -1.

Example

Input


5
2 1
0 1
3 2
0 1 0
5 2
1 1 0 1 0
4 2
0 1 0 1
1 1
0


Output


1
1
3
-1
0

Note

In the third sample test case the array will change as follows: 

  1. At the beginning a = [1, 1, 0, 1, 0], and a^{→ 2} = [1, 0, 1, 1, 0]. Their element-by-element "AND" is equal to $$$[1  \&  1, 1  \&  0, 0  \&  1, 1  \&  1, 0  \&  0] = [1, 0, 0, 1, 0]$$$ 
  2. Now a = [1, 0, 0, 1, 0], then a^{→ 2} = [1, 0, 1, 0, 0]. Their element-by-element "AND" equals to $$$[1  \&  1, 0  \&  0, 0  \&  1, 1  \&  0, 0  \&  0] = [1, 0, 0, 0, 0]$$$ 
  3. And finally, when a = [1, 0, 0, 0, 0] we get a^{→ 2} = [0, 0, 1, 0, 0]. Their element-by-element "AND" equals to $$$[1  \&  0, 0  \&  0, 0  \&  1, 0  \&  0, 0  \&  0] = [0, 0, 0, 0, 0]$$$ 

Thus, the answer is 3 steps.

In the fourth sample test case, the array will not change as it shifts by 2 to the right, so each element will be calculated as 0  \&  0 or 1  \&  1 thus not changing its value. So the answer is -1, the array will never contain only zeros.
#include <bits/stdc++.h>
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define drep(i,j,n) for(int i=0;i<(int)(n-1);i++)for(int j=i+1;j<(int)(n);j++)
#define trep(i,j,k,n) for(int i=0;i<(int)(n-2);i++)for(int j=i+1;j<(int)(n-1);j++)for(int k=j+1;k<(int)(n);k++)
#define codefor int test;scanf("%d",&test);while(test--)
#define INT(...) int __VA_ARGS__;in(__VA_ARGS__)
#define LL(...) ll __VA_ARGS__;in(__VA_ARGS__)
#define yes(ans) if(ans)printf("yes\n");else printf("no\n")
#define Yes(ans) if(ans)printf("Yes\n");else printf("No\n")
#define YES(ans) if(ans)printf("YES\n");else printf("NO\n")
#define popcount(v) __builtin_popcount(v)
#define vector1d(type,name,...) vector<type>name(__VA_ARGS__)
#define vector2d(type,name,h,...) vector<vector<type>>name(h,vector<type>(__VA_ARGS__))
#define vector3d(type,name,h,w,...) vector<vector<vector<type>>>name(h,vector<vector<type>>(w,vector<type>(__VA_ARGS__)))
using namespace std;
using ll = long long;
template<class T>using rpriority_queue = priority_queue<T, vector<T>, greater<T>>;
const int MOD=1000000007;
const int MOD2=998244353;
const int INF=1<<30;
const ll INF2=(ll)1<<60;
//入力系
void scan(int& a){scanf("%d",&a);}
void scan(long long& a){scanf("%lld",&a);}
template<class T,class L>void scan(pair<T, L>& p){scan(p.first);scan(p.second);}
template<class T> void scan(T& a){cin>>a;}
template<class T> void scan(vector<T>& vec){for(auto&& it:vec)scan(it);}
void in(){}
template <class Head, class... Tail> void in(Head& head, Tail&... tail){scan(head);in(tail...);}
//出力系
void print(const int& a){printf("%d",a);}
void print(const long long& a){printf("%lld",a);}
void print(const double& a){printf("%.15lf",a);}
template<class T,class L>void print(const pair<T, L>& p){print(p.first);putchar(' ');print(p.second);}
template<class T> void print(const T& a){cout<<a;}
template<class T> void print(const vector<T>& vec){if(vec.empty())return;print(vec[0]);for(auto it=vec.begin();++it!= vec.end();){putchar(' ');print(*it);}}
void out(){putchar('\n');}
template<class T> void out(const T& t){print(t);putchar('\n');}
template <class Head, class... Tail> void out(const Head& head,const Tail&... tail){print(head);putchar(' ');out(tail...);}
//デバッグ系
template<class T> void dprint(const T& a){cerr<<a;}
template<class T> void dprint(const vector<T>& vec){if(vec.empty())return;cerr<<vec[0];for(auto it=vec.begin();++it!= vec.end();){cerr<<" "<<*it;}}
void debug(){cerr<<endl;}
template<class T> void debug(const T& t){dprint(t);cerr<<endl;}
template <class Head, class... Tail> void debug(const Head& head, const Tail&... tail){dprint(head);cerr<<" ";debug(tail...);}
ll intpow(ll a, ll b){ ll ans = 1; while(b){ if(b & 1) ans *= a; a *= a; b /= 2; } return ans; }
ll modpow(ll a, ll b, ll p){ ll ans = 1; while(b){ if(b & 1) (ans *= a) %= p; (a *= a) %= p; b /= 2; } return ans; }
ll modinv(ll a, ll m) {ll b = m, u = 1, v = 0;while (b) {ll t = a / b;a -= t * b; swap(a, b);u -= t * v; swap(u, v);}u %= m;if (u < 0) u += m;return u;}
ll updivide(ll a,ll b){if(a%b==0) return a/b;else return (a/b)+1;}
template<class T> void chmax(T &a,const T b){if(b>a)a=b;}
template<class T> void chmin(T &a,const T b){if(b<a)a=b;}

namespace internal {
constexpr long long safe_mod(long long x, long long m) {x %= m;if (x < 0) x += m;return x;}
struct barrett {
    unsigned int _m;
    unsigned long long im;
    explicit barrett(unsigned int m) : _m(m), im((unsigned long long)(-1) / m + 1) {}
    unsigned int umod() const { return _m; }
    unsigned int mul(unsigned int a, unsigned int b) const {
        unsigned long long z = a;
        z *= b;
#ifdef _MSC_VER
        unsigned long long x;
        _umul128(z, im, &x);
#else
        unsigned long long x =
            (unsigned long long)(((unsigned long long)(z)*im));
#endif
        unsigned int v = (unsigned int)(z - x * _m);
        if (_m <= v) v += _m;
        return v;
    }
};
constexpr long long pow_mod_constexpr(long long x, long long n, int m) {
    if (m == 1) return 0;
    unsigned int _m = (unsigned int)(m);
    unsigned long long r = 1;
    unsigned long long y = safe_mod(x, m);
    while (n) {
        if (n & 1) r = (r * y) % _m;
        y = (y * y) % _m;
        n >>= 1;
    }
    return r;
}
constexpr bool is_prime_constexpr(int n) {
    if (n <= 1) return false;
    if (n == 2 || n == 7 || n == 61) return true;
    if (n % 2 == 0) return false;
    long long d = n - 1;
    while (d % 2 == 0) d /= 2;
    constexpr long long bases[3] = {2, 7, 61};
    for (long long a : bases) {
        long long t = d;
        long long y = pow_mod_constexpr(a, t, n);
        while (t != n - 1 && y != 1 && y != n - 1) {y = y * y % n;t <<= 1;}
        if (y != n - 1 && t % 2 == 0) {return false;}
    }
    return true;
}
template <int n> constexpr bool is_prime = is_prime_constexpr(n);
constexpr std::pair<long long, long long> inv_gcd(long long a, long long b) {
    a = safe_mod(a, b);
    if (a == 0) return {b, 0};
    long long s = b, t = a;
    long long m0 = 0, m1 = 1;
    while (t) {long long u = s / t;s -= t * u;m0 -= m1 * u;auto tmp = s;s = t;t = tmp;tmp = m0;m0 = m1;m1 = tmp;}
    if (m0 < 0) m0 += b / s;
    return {s, m0};
}
constexpr int primitive_root_constexpr(int m) {
    if (m == 2) return 1;
    if (m == 167772161) return 3;
    if (m == 469762049) return 3;
    if (m == 754974721) return 11;
    if (m == 998244353) return 3;
    int divs[20] = {};
    divs[0] = 2;
    int cnt = 1;
    int x = (m - 1) / 2;
    while (x % 2 == 0) x /= 2;
    for (int i = 3; (long long)(i)*i <= x; i += 2) {
        if (x % i == 0) {divs[cnt++] = i;while (x % i == 0) {x /= i;}}
    }
    if (x > 1) {divs[cnt++] = x;}
    for (int g = 2;; g++) {
        bool ok = true;
        for (int i = 0; i < cnt; i++) {
            if (pow_mod_constexpr(g, (m - 1) / divs[i], m) == 1) {ok = false;break;}
        }
        if (ok) return g;
    }
}
template <int m> constexpr int primitive_root = primitive_root_constexpr(m);
unsigned long long floor_sum_unsigned(unsigned long long n,unsigned long long m,unsigned long long a,unsigned long long b) {
    unsigned long long ans = 0;
    while (true) {
        if (a >= m) {ans += n * (n - 1) / 2 * (a / m);a %= m;}
        if (b >= m) {ans += n * (b / m);b %= m;}
        unsigned long long y_max = a * n + b;
        if (y_max < m) break;
        n = (unsigned long long)(y_max / m);
        b = (unsigned long long)(y_max % m);
        std::swap(m, a);
    }
    return ans;
}
template <class T> using is_integral = typename std::is_integral<T>;
template <class T>
using is_signed_int =
    typename std::conditional<is_integral<T>::value && std::is_signed<T>::value,
                              std::true_type,
                              std::false_type>::type;
template <class T>
using is_unsigned_int =
    typename std::conditional<is_integral<T>::value &&
                                  std::is_unsigned<T>::value,
                              std::true_type,
                              std::false_type>::type;
template <class T>
using to_unsigned = typename std::conditional<is_signed_int<T>::value,
                                              std::make_unsigned<T>,
                                              std::common_type<T>>::type;
template <class T> using is_signed_int_t = std::enable_if_t<is_signed_int<T>::value>;
template <class T> using is_unsigned_int_t = std::enable_if_t<is_unsigned_int<T>::value>;
template <class T> using to_unsigned_t = typename to_unsigned<T>::type;
struct modint_base {};
struct static_modint_base : modint_base {};
template <class T> using is_modint = std::is_base_of<modint_base, T>;
template <class T> using is_modint_t = std::enable_if_t<is_modint<T>::value>;
}  // namespace internal
template <int m, std::enable_if_t<(1 <= m)>* = nullptr>
struct static_modint : internal::static_modint_base {
    using mint = static_modint;
  public:
    static constexpr int mod() { return m; }
    static mint raw(int v) {mint x;x._v = v;return x;}
    static_modint() : _v(0) {}
    template <class T, internal::is_signed_int_t<T>* = nullptr>
    static_modint(T v) {
        long long x = (long long)(v % (long long)(umod()));
        if (x < 0) x += umod();_v = (unsigned int)(x);
    }
    template <class T, internal::is_unsigned_int_t<T>* = nullptr>
    static_modint(T v) {_v = (unsigned int)(v % umod());}
    unsigned int val() const { return _v; }
    mint& operator++() {_v++;if (_v == umod()) _v = 0;return *this;}
    mint& operator--() {if (_v == 0) _v = umod();_v--;return *this;}
    mint operator++(int) {mint result = *this;++*this;return result;}
    mint operator--(int) {mint result = *this;--*this;return result;}
    mint& operator+=(const mint& rhs) {_v += rhs._v;if (_v >= umod()) _v -= umod();return *this;}
    mint& operator-=(const mint& rhs) {_v -= rhs._v;if (_v >= umod()) _v += umod();return *this;}
    mint& operator*=(const mint& rhs) {unsigned long long z = _v;z *= rhs._v;_v = (unsigned int)(z % umod());return *this;}
    mint& operator/=(const mint& rhs) { return *this = *this * rhs.inv(); }
    mint operator+() const { return *this; }
    mint operator-() const { return mint() - *this; }
    mint pow(long long n) const {assert(0 <= n);mint x = *this, r = 1;while (n) {if (n & 1) r *= x;x *= x;n >>= 1;}return r;}
    mint inv() const {if (prime) {assert(_v);return pow(umod() - 2);} else {auto eg = internal::inv_gcd(_v, m);assert(eg.first == 1);return eg.second;}}
    friend mint operator+(const mint& lhs, const mint& rhs) {return mint(lhs) += rhs;}
    friend mint operator-(const mint& lhs, const mint& rhs) {return mint(lhs) -= rhs;}
    friend mint operator*(const mint& lhs, const mint& rhs) {return mint(lhs) *= rhs;}
    friend mint operator/(const mint& lhs, const mint& rhs) {return mint(lhs) /= rhs;}
    friend bool operator==(const mint& lhs, const mint& rhs) {return lhs._v == rhs._v;}
    friend bool operator!=(const mint& lhs, const mint& rhs) {return lhs._v != rhs._v;}
    //自分で加えたもの
    friend istream& operator>>(istream& os,mint& rhs) noexcept {
        long long v;
        rhs = mint{(os >> v, v)};
        return os;
    }
    friend constexpr ostream& operator << (ostream &os, const mint& rhs) noexcept {
        return os << rhs._v;
    }
  private:
    unsigned int _v;
    static constexpr unsigned int umod() { return m; }
    static constexpr bool prime = internal::is_prime<m>;
};
using mint=static_modint<1000000007>;
using mint2=static_modint<998244353>;

vector<ll> dijkstra(int start,vector<vector<pair<int,ll>>> &g){
    int n=g.size();
    vector<ll> dist(n,INF2);
    vector<bool> check(n,false);
    rpriority_queue<pair<ll,int>> pq;
    dist[start]=0;
    pq.push({0,0});
    ll d,c;
    int v,u;
    while(!pq.empty()){
        tie(d,v)=pq.top();
        pq.pop();
        if(check[v])continue;
        check[v]=true;
        for(auto p:g[v]){
            tie(u,c)=p;
            if(d+c>=dist[u])continue;
            dist[u]=d+c;
            pq.push({d+c,u});
        }
    }
    return dist;
}

int main(){
    codefor{
        INT(n,d);
        vector<int> a(n);
        in(a);
        int v=__gcd(n,d);
        vector<vector<int>> g(v);
        vector<int> t(v,1);
        rep(i,n){
            g[i%v].push_back(i);
            t[i%v]&=a[i];
        }
        bool D=true;
        rep(i,v){
            if(t[i]){
                D=false;
                break;
            }
        }
        if(!D){
            out(-1);
            continue;
        }
        int ans=0;
        rep(i,v){
            int m=n/v;
            vector<int> dist(m,INF);
            queue<int> next;
            rep(j,g[i].size()){
                if(a[g[i][j]]==0){
                    dist[g[i][j]/v]=0;
                    next.push(g[i][j]/v);
                }
            }
            while(!next.empty()){
                int u=next.front();
                next.pop();
                int to=((u+d)/v+m)%m;
                if(dist[u]+1<dist[to]){
                    dist[to]=dist[u]+1;
                    next.push(to);
                    chmax(ans,dist[to]);
                }
            }
            //debug("{",dist);
        }
        out(ans);
    }
}