#pragma GCC optimize(3,"Ofast","inline")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define pb push_back
#define fi first
#define se second
//#define int long long
const int N = 5e5 + 5;
int a[N], t[N], cnt = 0, len;

struct node {
	int ls, rs, cnt = 0;
	ll sum = 0;
} tr[N * 20];
int id = 0;

int updata(int p, int pre, int l, int r, int pos, int k) {
	p = ++id;
	//cout << id << '\n';
	tr[p]=tr[pre];
	tr[p].sum += k;
	tr[p].cnt += 1;
	//cout << l << ' ' << r << ' ' << tr[p].sum << ' ' << tr[p].cnt << '\n';
	if (l == r) {
		return p;
	}
	int mid = (l + r) / 2;
	if (pos <= mid)
		tr[p].ls = updata(tr[p].ls, tr[pre].ls, l, mid, pos, k);
	else
		tr[p].rs = updata(tr[p].rs, tr[pre].rs, mid + 1, r, pos, k);
	return p;
}
ll ct,st,ql,qr;
void get(int p, int l, int r) {
	if (ql <= l && r <= qr) {
		ct+=tr[p].cnt;
		st+=tr[p].sum;
		return ;
	}
	int mid = (l + r) / 2;
	if (mid >= ql) {
		get(tr[p].ls, l, mid);
	}
	if (mid < qr) {
		get(tr[p].rs, mid + 1, r);
	}
}
int rt[N];

pair<int, ll> query1(int l, int r, int x) {	
	st=ct=0;
	ql=x+1,qr=len;
	get(rt[r], 1, len);
	ll c1=ct,s1=st;
	st=ct=0;
	get(rt[l - 1], 1, len);
	return {c1 - ct, s1 - st};
}

int query2(int p1, int p2, int l, int r, int k) {
	if (l == r)
		return l;
	int mid = (l + r) / 2;
	int c = tr[tr[p1].ls].cnt - tr[tr[p2].ls].cnt;
	if (c >= k)
		return query2(tr[p1].ls, tr[p2].ls, l, mid, k);
	return query2(tr[p1].rs, tr[p2].rs, mid + 1, r, k - c);
}

int query2(int l, int r) {
	int mid = (r - l + 2) / 2;
	return query2(rt[r], rt[l - 1], 1, len, mid);
}
int pos[N];
ll pre[N];

void solve() {
	ll n, k;
	cin >> n >> k;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
		a[i] -= i;
		t[++cnt] = a[i];
		pre[i] = pre[i - 1] + a[i];
	}
	sort(t + 1, t + 1 + cnt);
	len = unique(t + 1, t + 1 + cnt) - t - 1;
	for (int i = 1; i <= n; i++) {
		//cout << a[i] << ' ';
		pos[i] = lower_bound(t + 1, t + 1 + len, a[i]) - t;
		//cout << pos[i] << ' ';
		//continue;
		rt[i] = updata(rt[i], rt[i - 1], 1, len, pos[i], a[i]);
	}

	auto check = [&](int x) {
		for (int i = x ; i <= n ; i++) {
			int l = i - x + 1, r = i;
			int zh = query2(i - x + 1, i);
			ll xt = t[zh];
			auto [c1, s1] = query1(l, r, zh);
			ll c2 = r - l + 1 - c1, s2 = pre[r] - pre[l - 1] - s1;
			ll ans = c2 * xt - s2 + (s1 - c1 * xt);

			if (ans <= k)
				return true;
		}
		return false;
	};

	int le = 1, ri = n, ans = 1;
	while (le <= ri) {
		int mid = (le + ri) >> 1;
		//cout << ri << ' ' << mid << ' ' << le << '\n';
		if (check(mid)) {
			le = mid + 1;
			ans = mid;
		} else
			ri = mid - 1;
	}

	cout << ans << '\n';
	cnt = 0;
	id = 0;
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int t = 1;
	cin >> t;
	while (t--)
		solve();
	return 0;
}