1. #include <cstdio>
    2. #include <cstring>
    3. #include <cmath>
    4. #include <cstdlib>
    5. #include <iostream>
    6. #include <cassert>
    7. using namespace std;
    8. inline int gi(){
    9.     char tmp=getchar();int ans=0;
    10.     while(!isdigit(tmp)) tmp=getchar();
    11.     while(isdigit(tmp)){
    12.         ans = ans * 10 + tmp - '0';
    13.         tmp = getchar();
    14.     }
    15.     return ans;
    16. }
    17. const int N = 230;
    18. double P[N];int D[N];
    19. double Pow[N][332],F[N][N];
    20. double G[N];
    21. signed main()
    22. {
    23. #ifdef TSUKIAKIOI
    24.     freopen("data.in","r",stdin);
    25. #endif
    26.     int T=gi();
    27.     while(T--){
    28.         int n,r;n=gi(),r=gi();
    29.         memset(P,0,sizeof P);
    30.         memset(Pow,0,sizeof P);
    31.         memset(F,0,sizeof F);
    32.         memset(G,0,sizeof G);
    33.         for(int i=1;i<=n;++i){
    34.             scanf("%lf",&P[i]);
    35.             D[i] = gi();
    36.         }
    37.         for(int i=1;i<=n;++i){
    38.             Pow[i][0] = 1;
    39.             for(int j=1;j<=r;++j)
    40.                 Pow[i][j] = Pow[i][j-1] * (1.0-P[i]);
    41.         }
    42.         G[1] = 1.0 - Pow[1][r];
    43.         F[1][0] = Pow[1][r];
    44.         F[1][1] = 1-F[1][0];
    45.         for(int i=2;i<=n;++i){
    46.             for(int j=0;j<=min(r,i);++j)
    47.                 if(!j)
    48.                     F[i][j] = Pow[i][r-j] * F[i-1][j];
    50.                 else if(j)
    51.                     F[i][j] = 
    52.                     (1.0-Pow[i][r-j+1]) * F[i-1][j-1] +
    53.                     (Pow[i][r-j]) * F[i-1][j];
    54.         }
    55.         for(int i=2;i<=n;++i){
    56.             double res = 0;
    57.             for(int j=0;j<=min(i,r);++j)
    58.                 res += F[i-1][j] * (1.0-Pow[i][r-j]);
    59.             G[i] = res;
    60.         }
    61.         double ans = 0;
    62.         for(int i=1;i<=n;++i)
    63.             ans += 1.0*G[i]*D[i];
    64.         printf("%.10lf\n",ans);
    65.     }
    66.     return 0;
    67. }