// Problem: Shopping
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/20960/1037
// Memory Limit: 1048576 MB
// Time Limit: 2000 ms
// 2022-03-04 20:13:26
//
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
ll ans = 1;
while (b) {
if (b & 1) ans = ans * a%mod;
a = a * a %mod;
b >>= 1;
}
return ans;
}
inline int read () {
int x = 0, f = 0;
char ch = getchar();
while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
return f?-x:x;
}
template<typename T> void print(T x) {
if (x < 0) putchar('-'), x = -x;
if (x >= 10) print(x/10);
putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
return (a + b) %mod;
}
inline ll inv (ll a) {
return qmi(a, mod - 2);
}
const int N = 1e3 + 10;
int n, m;
vector<int> a, b;
void solve() {
cin >> n>> m;
int f = 0;
a.clear();
for (int i = 1; i <= n; i ++) {
int u, v;
cin >> u >> v;
a.pb(u);
f += v;
}
double ans = 0;
sort(all(a));

for (int i = n - min(f, m); i < n; i ++)
ans += a[i]*1.0 / 2;
for (int i = 0; i < n - min(f, m); i ++)
ans += a[i] ;
printf("%.1lf\n", ans);

}
int main () {
int t;
t =1;
cin >> t;
while (t --) solve();
return 0;
}


思考这个问题的时候,是想着一个购物购物的看,但是这样就会陷入对将什么样的武平放入购物篮的思考。实际上这里只需要从小到大排序,m和f的大小决定了多少个物品能半价买