#include <bits/stdc++.h>
    #include <chrono>
    using namespace std;
    using namespace chrono;
    // "AJEET JAIN"----"JAI JINENDRA"  
    /* "णमो अरिहंताणं",
        "णमो सिद्धाणं",
        "णमो आयरियाणं",
        "णमो उवज्झायाणं",
        "णमो लोए सव्वसाहूणं",
        "",
        "एसो पंच नमोक्कारो, सव्व पावप्पणासणो",
        "मंगलाणं च सव्वेसिं, पडमं हवै मंगलं",   */
 
 
    // Aliases to op
    using ll = long long;
    using ull = unsigned long long;
    using ld = double;
    using vll = vector<ll>;
 
 
    // Constants
    constexpr ll INF = 4e18;
    constexpr ld EPS = 1e-9;
    constexpr ll MOD = 1e9 + 7;
 
 
 
    // Macros
    #define F first
    #define S second
    #define all(x) begin(x), end(x)
    #define allr(x) rbegin(x), rend(x)
    #define py cout<<"YES\n";
    #define pn cout<<"NO\n";
    #define forn(i,n) for(int i=0;i<n;i++)
    #define for1(i,n) for(int i=1;i<=n;i++)
 
    // #define insert push_back
    #define pb push_back
    #define MP make_pair
    #define endl '\n'
 
    /*
      remove substring or subarray ---> try to think about sliding w
    
    */                  
 
     /*
      
     Golden Rule
 
     1) problem is easy
     2) proofs is easy
     3) implementation is easy
     
     /*
         ROUGH --
                 
     */
    long long countSubarraysWithSumAtMostK(vector<char> &b, long long k) {
        int n = b.size();
        long long i = 0, diff = 0, count = 0 , ans = 0;
        multiset<char> m;
 
        for (int j = 0; j < n; j++) {
           m.insert(b[j]);
 
            while (*m.rbegin() - *m.begin() > k && !m.empty()) {
                m.erase(m.find(b[i]));
                i++;
            }
            count = (j - i + 1);
            ans = max( ans , count );
        }
 
        return ans;
    }
 
 
    void AJNJ(){
       ll n,k;
       cin>>n>>k;
 
       vector<char> b(n);
       for1(i , n){
          cin >> b[i];
       }
 
       cout << countSubarraysWithSumAtMostK(b , k) << endl;
 
    }
 
 
    int main(){
        ios::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
        int T = 1;
        cin>>T;
        auto start1 = high_resolution_clock::now();
        while(T--){
            AJNJ();
        }
        auto stop1 = high_resolution_clock::now();
        auto duration = duration_cast<microseconds>(stop1 - start1);
        cerr << "Time: " << duration . count() / 1000 << " ms" << endl;
 
        return 0;
    }

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