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: Time Limit Exceeded)

use std::cmp::Reverse;
use std::collections::{BinaryHeap, HashSet};
#[derive(Default)]
//NOTE
//declare variables to reduce the number of parameters for dp and dfs etc.
struct Solver {}

type MinHeap<T> = BinaryHeap<Reverse<T>>;

impl Solver {
    fn solve(&mut self) {
        let stdin = std::io::stdin();
        let mut scn = Scanner {
            stdin: stdin.lock(),
        };
        let n: usize = scn.read();
        let m: usize = scn.read();
        let s: usize = scn.read();
        let mut edges = vec![vec![]; n];
        (0..m).for_each(|_| {
            let u: usize = scn.read();
            let v: usize = scn.read();
            let a: usize = scn.read();
            let b: usize = scn.read();
            edges[u - 1].push((v - 1, a, b));
            edges[v - 1].push((u - 1, a, b));
        });
        let cd: Vec<_> = (0..n)
            .map(|_| {
                let c: usize = scn.read();
                let d: usize = scn.read();
                (c, d)
            })
            .collect();
        let mut min_heap = MinHeap::new();
        min_heap.push(Reverse((0, 0, std::cmp::min(s, n * 50), 0)));
        let mut visited = vec![vec![vec![false; 2501]; 120];50];
        let mut costs = vec![None; n];

        while let Some(p) = min_heap.pop() {
            let (cost, pos, coin, cnt) = p.0;

            if visited[pos][cnt][coin] {
                continue;
            }

            if cnt > 2 * n +  1{
                continue;
            }
            visited[pos][cnt][coin] = true;

            if let Some(c) = costs[pos] {
                if cost < c {
                    costs[pos] = Some(cost);
                }
            } else {
                costs[pos] = Some(cost);
            }

            for i in 0..=n * 50 {
                let coin = coin + i * cd[pos].0;
                let coin_cost = i * cd[pos].1;
                if coin > n * 50 {
                    break;
                }

                for &nxt in edges[pos].iter() {
                    let (nxt, a, b) = nxt;
                    if coin >= a {
                        if !visited[nxt][cnt + 1][coin - a] {
                            min_heap.push(Reverse((cost + b + coin_cost, nxt, coin - a, cnt + 1)));
                        }
                    }
                }
            }
        }
        for x in costs.iter().skip(1) {
            println!("{}", x.unwrap());
        }
    }
}

fn main() {
    std::thread::Builder::new()
        .stack_size(64 * 1024 * 1024) // 64MB
        .spawn(|| {
            let mut solver = Solver::default();
            solver.solve()
        })
        .unwrap()
        .join()
        .unwrap();
}
pub mod utils {
    const DYX: [(isize, isize); 8] = [
        (0, 1),   //right
        (1, 0),   //down
        (0, -1),  //left
        (-1, 0),  //top
        (1, 1),   //down right
        (-1, 1),  //top right
        (1, -1),  //down left
        (-1, -1), //top left
    ];

    pub fn try_adj(y: usize, x: usize, dir: usize, h: usize, w: usize) -> Option<(usize, usize)> {
        let ny = y as isize + DYX[dir].0;
        let nx = x as isize + DYX[dir].1;
        if ny >= 0 && nx >= 0 {
            let ny = ny as usize;
            let nx = nx as usize;
            if ny < h && nx < w {
                Some((ny, nx))
            } else {
                None
            }
        } else {
            None
        }
    }
}

//snippet from kenkoooo
pub struct Scanner<R> {
    stdin: R,
}

impl<R: std::io::Read> Scanner<R> {
    pub fn read<T: std::str::FromStr>(&mut self) -> T {
        use std::io::Read;
        let buf = self
            .stdin
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b' ' || b == b'\n' || b == b'\r')
            .take_while(|&b| b != b' ' && b != b'\n' && b != b'\r')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn read_line(&mut self) -> String {
        use std::io::Read;
        let buf = self
            .stdin
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b'\n' || b == b'\r')
            .take_while(|&b| b != b'\n' && b != b'\r')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }

    pub fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}