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You live in a small town with *R* bidirectional roads connecting *C* crossings and you want to go from crossing 1 to crossing *C* as soon as possible. You can visit other crossings before arriving at crossing C, but it’s not mandatory.

You have exactly one chance to ask your friend to repair exactly one existing road, __from the time you __**leave crossing 1**. If he repairs the

You cannot start to cross the road when your friend is repairing it.

There will be at most 25 test cases. Each test case begins with two integers C and R (2<=C<=100, 1<=R<=500). Each of the next R lines contains two integers xi, yi (1<=xi, yi<=C) and two positive floating-point numbers vi and ai (1<=vi<=20,1<=ai<=5), indicating that there is a bidirectional road connecting crossing xi and yi, with parameters vi and ai (see above). Each pair of crossings can be connected by at most one road. The input is terminated by a test case with C=R=0, you should not process it.

For each test case, print the smallest time it takes to reach crossing C from crossing 1, rounded to 3 digitsafter decimal point. It’s always possible to reach crossing C from crossing 1.

```
3 2
1 2 1.5 1.8
2 3 2.0 1.5
2 1
1 2 2.0 1.8
0 0
```

```
2.589
1.976
```