#define _CRT_SECURE_NO_WARNINGS 1
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<set>
#include<queue>
#define x first
#define y second
using namespace std;
const double PI = acos(-1);
const int N = 5e5 + 10;
typedef long long ll;
typedef pair<ll, ll> PII;
int n;
ll a[N], k;
struct MID {
    multiset<ll, greater<ll>>p;//存储较小部分，begin为最大值
    multiset<ll>q;//存储较大部分，begin为最小值
    ll sump, sumq;
    MID() {
        sump = sumq = 0;
    }
    void move() {
        if (abs((int)p.size() - (int)q.size()) > 1) {
            if (p.size() > q.size()) {
                auto t = *p.begin();
                p.extract(t), q.insert(t);
                sump -= t, sumq += t;
            }
            else {
                auto t = *q.begin();
                q.extract(t), p.insert(t);
                sumq -= t, sump += t;
            }
        }
    }
    void add(ll x) {
        if (!p.size() || x <= *p.begin()) {
            p.insert(x);
            sump += x;
        }
        else {
            q.insert(x);
            sumq += x;
        }
        move();
    }
    void del(ll x) {
        if (p.count(x)) {
            p.extract(x);
            sump -= x;
        }
        else {
            q.extract(x);
            sumq -= x;
        }
        move();
    }
    ll get() {
        ll mid = 0;
        if (p.size() && p.size() >= q.size())mid = *p.begin();
        else if (q.size())mid = *q.begin();
        return mid * p.size() - sump + sumq - mid * q.size();
    }
    ll getmid() {
        ll mid = 0;
        if (p.size() && p.size() >= q.size())mid = *p.begin();
        else if (q.size())mid = *q.begin();
        return mid;
    }
    int size() {
        return p.size() + q.size();
    }
};
int main() {
    int _ = 1;
    scanf("%d", &_);
    while (_--) {
        scanf("%d%lld", &n, &k);
        for (int i = 1; i <= n; i++) {
            scanf("%lld", &a[i]);
            a[i] -= i;
        }
        //for (int i = 1; i <= n; i++)cout << a[i] << " ";
        //cout << '\n';
        MID s;
        int ans = 1;
        for (int r = 0, l = 1; l <= n; l++) {
            while (r < n) {
                s.add(a[++r]);
                if (s.get() > k) {
                    s.del(a[r--]);
                    break;
                }
            }
            ans = max(ans, r - l + 1);
            s.del(a[l]);
        }
        printf("%d\n", ans);
    }
    return 0;
}