codeforces865C

Gotta Go Fast

 CodeForces - 865C 

You're trying to set the record on your favorite video game. The game consists of Nlevels, which must be completed sequentially in order to beat the game. You usually complete each level as fast as possible, but sometimes finish a level slower. Specifically, you will complete the i-th level in either F_i seconds or S_i seconds, where F_i < S_i, and there's a P_i percent chance of completing it in F_i seconds. After completing a level, you may decide to either continue the game and play the next level, or reset the game and start again from the first level. Both the decision and the action are instant.

Your goal is to complete all the levels sequentially in at most R total seconds. You want to minimize the expected amount of time playing before achieving that goal. If you continue and reset optimally, how much total time can you expect to spend playing?

Input

The first line of input contains integers N and R , the number of levels and number of seconds you want to complete the game in, respectively. N lines follow. The ith such line contains integers F_i, S_i, P_i (1 ≤ F_i < S_i ≤ 100, 80 ≤ P_i ≤ 99), the fast time for level i, the slow time for level i, and the probability (as a percentage) of completing level i with the fast time.

Output

Print the total expected time. Your answer must be correct within an absolute or relative error of 10^- 9.

Formally, let your answer be a, and the jury's answer be b. Your answer will be considered correct, if .

Examples

Input

1 8
2 8 81

Output

3.14

Input

2 30
20 30 80
3 9 85

Output

31.4

Input

4 319
63 79 89
79 97 91
75 87 88
75 90 83

Output

314.159265358

Note

In the first example, you never need to reset. There's an 81% chance of completing the level in 2 seconds and a 19% chance of needing 8 seconds, both of which are within the goal time. The expected time is 0.81·2 + 0.19·8 = 3.14.

In the second example, you should reset after the first level if you complete it slowly. On average it will take 0.25 slow attempts before your first fast attempt. Then it doesn't matter whether you complete the second level fast or slow. The expected time is 0.25·30 + 20 + 0.85·3 + 0.15·9 = 31.4.

sol：

dp[i][j]表示当前为第i关，已用时j，从当前开始通关的用时期望
tmp表示从头开始通关的用时期望
设当前状态为(i,j)：
①如果在挑战第i关前选择重新开始游戏，则通关的期望值tmp
②如果通过第i关用时为a[i],则继续进行游戏并通关的期望值为(dp[i+1][j+a[i]]+a[i])*p[i]
③如果通过第i关用时为b[i],则继续进行游戏并通关的期望值为(dp[i+1][j+b[i]]+b[i])*(1-p[i])

/*
dp[i][j]表示当前为第i关，已用时j，从当前开始通关的用时期望
tmp表示从头开始通关的用时期望
设当前状态为(i,j)：
①如果在挑战第i关前选择重新开始游戏，则通关的期望值tmp
②如果通过第i关用时为a[i],则继续进行游戏并通关的期望值为(dp[i+1][j+a[i]]+a[i])*p[i]
③如果通过第i关用时为b[i],则继续进行游戏并通关的期望值为(dp[i+1][j+b[i]]+b[i])*(1-p[i])
*/
#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=0; bool f=0; char ch=' ';
    while(!isdigit(ch)) {f|=(ch=='-'); ch=getchar();}
    while(isdigit(ch)) {s=(s<<3)+(s<<1)+(ch^48); ch=getchar();}
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0) {putchar('-'); x=-x;}
    if(x<10) {putchar(x+'0'); return;}
    write(x/10); putchar((x%10)+'0');
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('
')
const int N=55,M=5505;
const double eps=1e-8;
int n,m,a[N],b[N],p[N];
double dp[N][M];
inline bool chk(double tmp)
{
    int i,j;
    for(i=n;i>=1;i--)
    {
        for(j=m+1;j<M;j++) dp[i+1][j]=tmp; //重来
        for(j=0;j<=m;j++)
        {
            double t1=(double)(dp[i+1][j+a[i]]+a[i])*p[i]/100;
            double t2=(double)(dp[i+1][j+b[i]]+b[i])*(100-p[i])/100;
            dp[i][j]=min(t1+t2,tmp);
        }
    }
    return dp[1][0]<tmp;
}
int main()
{
    freopen("data.in","r",stdin);
    int i;
    R(n); R(m);
    for(i=1;i<=n;i++)
    {
        R(a[i]); R(b[i]); R(p[i]);
    }
    double l=0.00,r=1e10,mid;
    for(i=1;i<=100;i++)
    {
        mid=(l+r)*0.50;
        if(chk(mid)) r=mid;
        else l=mid;
    }
    printf("%.12lf
",l);
    return 0;
}
/*
input
4 319
63 79 89
79 97 91
75 87 88
75 90 83
output
314.159265358
*/