Problem:
Score : 500 points

Problem Statement

There are N cities numbered 1 to N, connected by M railroads.

You are now at City 1, with 10^{100} gold coins and S silver coins in your pocket.

The i-th railroad connects City U_i and City V_i bidirectionally, and a one-way trip costs A_i silver coins and takes B_i minutes.
You cannot use gold coins to pay the fare.

There is an exchange counter in each city. At the exchange counter in City i, you can get C_i silver coins for 1 gold coin.
The transaction takes D_i minutes for each gold coin you give.
You can exchange any number of gold coins at each exchange counter.

For each t=2, ..., N, find the minimum time needed to travel from City 1 to City t. You can ignore the time spent waiting for trains.

Constraints

2 \leq N \leq 50

N-1 \leq M \leq 100

0 \leq S \leq 10^9

1 \leq A_i \leq 50

1 \leq B_i,C_i,D_i \leq 10^9

1 \leq U_i < V_i \leq N

There is no pair i, j(i \neq j) such that (U_i,V_i)=(U_j,V_j).

Each city t=2,...,N can be reached from City 1 with some number of railroads.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M S
U_1 V_1 A_1 B_1
:
U_M V_M A_M B_M
C_1 D_1
:
C_N D_N

Output

For each t=2, ..., N in this order, print a line containing the minimum time needed to travel from City 1 to City t.

Sample Input 1

3 2 1
1 2 1 2
1 3 2 4
1 11
1 2
2 5

Sample Output 1

2
14

The railway network in this input is shown in the figure below.

In this figure, each city is labeled as follows:

The first line: the ID number i of the city (i for City i)

The second line: C_i / D_i

Similarly, each railroad is labeled as follows:

The first line: the ID number i of the railroad (i for the i-th railroad in input)

The second line: A_i / B_i

You can travel from City 1 to City 2 in 2 minutes, as follows:

Use the 1-st railroad to move from City 1 to City 2 in 2 minutes.

You can travel from City 1 to City 3 in 14 minutes, as follows:

Use the 1-st railroad to move from City 1 to City 2 in 2 minutes.

At the exchange counter in City 2, exchange 3 gold coins for 3 silver coins in 6 minutes.

Use the 1-st railroad to move from City 2 to City 1 in 2 minutes.

Use the 2-nd railroad to move from City 1 to City 3 in 4 minutes.

Sample Input 2

4 4 1
1 2 1 5
1 3 4 4
2 4 2 2
3 4 1 1
3 1
3 1
5 2
6 4

Sample Output 2

5
5
7

The railway network in this input is shown in the figure below:

You can travel from City 1 to City 4 in 7 minutes, as follows:

At the exchange counter in City 1, exchange 2 gold coins for 6 silver coins in 2 minutes.

Use the 2-nd railroad to move from City 1 to City 3 in 4 minutes.

Use the 4-th railroad to move from City 3 to City 4 in 1 minutes.

Sample Input 3

6 5 1
1 2 1 1
1 3 2 1
2 4 5 1
3 5 11 1
1 6 50 1
1 10000
1 3000
1 700
1 100
1 1
100 1

Sample Output 3

1
9003
14606
16510
16576

The railway network in this input is shown in the figure below:

You can travel from City 1 to City 6 in 16576 minutes, as follows:

Use the 1-st railroad to move from City 1 to City 2 in 1 minute.

At the exchange counter in City 2, exchange 3 gold coins for 3 silver coins in 9000 minutes.

Use the 1-st railroad to move from City 2 to City 1 in 1 minute.

Use the 2-nd railroad to move from City 1 to City 3 in 1 minute.

At the exchange counter in City 3, exchange 8 gold coins for 8 silver coins in 5600 minutes.

Use the 2-nd railroad to move from City 3 to City 1 in 1 minute.

Use the 1-st railroad to move from City 1 to City 2 in 1 minute.

Use the 3-rd railroad to move from City 2 to City 4 in 1 minute.

At the exchange counter in City 4, exchange 19 gold coins for 19 silver coins in 1900 minutes.

Use the 3-rd railroad to move from City 4 to City 2 in 1 minute.

Use the 1-st railroad to move from City 2 to City 1 in 1 minute.

Use the 2-nd railroad to move from City 1 to City 3 in 1 minute.

Use the 4-th railroad to move from City 3 to City 5 in 1 minute.

At the exchange counter in City 5, exchange 63 gold coins for 63 silver coins in 63 minutes.

Use the 4-th railroad to move from City 5 to City 3 in 1 minute.

Use the 2-nd railroad to move from City 3 to City 1 in 1 minute.

Use the 5-th railroad to move from City 1 to City 6 in 1 minute.

Sample Input 4

4 6 1000000000
1 2 50 1
1 3 50 5
1 4 50 7
2 3 50 2
2 4 50 4
3 4 50 3
10 2
4 4
5 5
7 7

Sample Output 4

1
3
5

The railway network in this input is shown in the figure below:

Sample Input 5

2 1 0
1 2 1 1
1 1000000000
1 1

Sample Output 5

1000000001

The railway network in this input is shown in the figure below:

You can travel from City 1 to City 2 in 1000000001 minutes, as follows:

At the exchange counter in City 1, exchange 1 gold coin for 1 silver coin in 1000000000 minutes.

Use the 1-st railroad to move from City 1 to City 2 in 1 minute.

Solution in Rust (1.42.0) (judge verdict: Accepted)

use std::cmp::min;
use std::cmp::Reverse;
use std::io::{stdin, Read};

macro_rules! parse {
    ($it: ident ) => {};
    ($it: ident, ) => {};
    ($it: ident, $var:ident : $t:tt $($r:tt)*) => {
        let $var = parse_val!($it, $t);
        parse!($it $($r)*);
    };
    ($it: ident, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = parse_val!($it, $t);
        parse!($it $($r)*);
    };
}

macro_rules! parse_val {
    ($it: ident, [$t:tt; $len:expr]) => {
        (0..$len).map(|_| parse_val!($it, $t)).collect::<Vec<_>>();
    };
    ($it: ident, ($($t: ty),*)) => {
        ($(parse_val!($it, $t)),*)
    };
    ($it: ident, $t: ty) => {
        $it.next().unwrap().parse::<$t>().unwrap()
    };
}

fn solve(s: &str) {
    let mut it = s.split_whitespace();
    parse!(it, n: usize, m: usize, s: usize);
    let mut g = vec![vec![]; n];
    let mut cd = vec![(0, 0); n];
    let mut maxa = 0;
    for _ in 0..m {
        parse!(it, u: usize, v: usize, a: usize, b: usize);
        g[u - 1].push((v - 1, a, b));
        g[v - 1].push((u - 1, a, b));
        maxa = std::cmp::max(maxa, a);
    }
    for i in 0..n {
        parse!(it, c: usize, d: usize);
        cd[i] = (c, d);
    }
    let mut q = std::collections::BinaryHeap::new();
    let maxs: usize = maxa * n;
    let mut dp = vec![vec![std::usize::MAX / 100; maxs + 1]; n];
    let mut marks = vec![vec![false; maxs + 1]; n];
    let s0 = if s >= maxs { maxs } else { s };
    dp[0][s0] = 0;
    q.push(Reverse((0, (0, s0))));
    while let Some(Reverse((l, (u, s)))) = q.pop() {
        if marks[u][s] {
            continue;
        }
        dp[u][s] = l;
        marks[u][s] = true;
        let (c, d) = cd[u];
        q.push(Reverse((
            l + d,
            (u, if s + c >= maxs { maxs } else { s + c }),
        )));
        for &(v, a, b) in g[u].iter() {
            if s >= a && !marks[v][s - a] {
                q.push(Reverse((l + b, (v, s - a))))
            }
        }
    }
    for i in 1..n {
        println!("{}", *dp[i].iter().min().unwrap());
    }
}

fn main() {
    let mut s = String::new();
    stdin().read_to_string(&mut s).unwrap();
    solve(&s);
}

#[cfg(test)]
mod tests {
    use super::*;
    #[test]
    fn test_input() {
        let s = "
        ";
        solve(s);
    }
}