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)

#![allow(non_snake_case)]
#![allow(dead_code)]
#![allow(unused_macros)]
#![allow(unused_imports)]

use std::str::FromStr;
use std::io::*;
use std::collections::*;
use std::cmp::*;

struct Scanner<I: Iterator<Item = char>> {
    iter: std::iter::Peekable<I>,
}

macro_rules! exit {
    () => {{
        exit!(0)
    }};
    ($code:expr) => {{
        if cfg!(local) {
            writeln!(std::io::stderr(), "===== Terminated =====")
                .expect("failed printing to stderr");
        }
        std::process::exit($code);
    }}
}

impl<I: Iterator<Item = char>> Scanner<I> {
    pub fn new(iter: I) -> Scanner<I> {
        Scanner {
            iter: iter.peekable(),
        }
    }

    pub fn safe_get_token(&mut self) -> Option<String> {
        let token = self.iter
            .by_ref()
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect::<String>();
        if token.is_empty() {
            None
        } else {
            Some(token)
        }
    }

    pub fn token(&mut self) -> String {
        self.safe_get_token().unwrap_or_else(|| exit!())
    }

    pub fn get<T: FromStr>(&mut self) -> T {
        self.token().parse::<T>().unwrap_or_else(|_| exit!())
    }

    pub fn vec<T: FromStr>(&mut self, len: usize) -> Vec<T> {
        (0..len).map(|_| self.get()).collect()
    }

    pub fn mat<T: FromStr>(&mut self, row: usize, col: usize) -> Vec<Vec<T>> {
        (0..row).map(|_| self.vec(col)).collect()
    }

    pub fn char(&mut self) -> char {
        self.iter.next().unwrap_or_else(|| exit!())
    }

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

    pub fn mat_chars(&mut self, row: usize) -> Vec<Vec<char>> {
        (0..row).map(|_| self.chars()).collect()
    }

    pub fn line(&mut self) -> String {
        if self.peek().is_some() {
            self.iter
                .by_ref()
                .take_while(|&c| !(c == '\n' || c == '\r'))
                .collect::<String>()
        } else {
            exit!();
        }
    }

    pub fn peek(&mut self) -> Option<&char> {
        self.iter.peek()
    }
}

#[derive(Clone)]
struct Edge(usize, usize, usize);

const INF: usize = 1 << 60;
const MAX: usize = 50;

fn main() {
    let cin = stdin();
    let cin = cin.lock();
    let mut sc = Scanner::new(cin.bytes().map(|c| c.unwrap() as char));
    let N: usize = sc.get();
    let M: usize = sc.get();
    let S: usize = sc.get();
    let mut adj = vec![Vec::new(); N];
    for _ in 0..M {
        let U: usize = sc.get::<usize>()-1;
        let V: usize = sc.get::<usize>()-1;
        let A: usize = sc.get();
        let B: usize = sc.get();
        adj[U].push(Edge(V, A, B));
        adj[V].push(Edge(U, A, B));
    }
    let mut exchange = Vec::new();
    for _ in 0..N {
        let C: usize = sc.get();
        let D: usize = sc.get();
        exchange.push((C, D));
    }
    let mut dp = vec![vec![INF; MAX * MAX + 1]; N];
    dp[0][min(MAX * MAX, S)] = 0;
    let mut bh = BinaryHeap::new();
    bh.push((Reverse(0usize), 0usize, min(MAX * MAX, S)));
    while let Some((Reverse(cost), now, coin)) = bh.pop() {
        if dp[now][coin] < cost {
            continue;
        }
        for &Edge(to, fare, time) in &adj[now] {
            if coin >= fare {
                if dp[to][coin - fare] > dp[now][coin] + time {
                    dp[to][coin - fare] = dp[now][coin] + time;
                    bh.push((Reverse(cost + time), to, coin - fare));
                }
            }
        }
        let (C, D) = exchange[now];
        if coin + C <= MAX * MAX && dp[now][coin + C] > dp[now][coin] + D {
            dp[now][coin + C] = min(dp[now][coin + C], dp[now][coin] + D);
            bh.push((Reverse(cost + D), now, coin + C));
        }
    }
    for t in 1..N {
        let mut ans = INF;
        for c in 0..=MAX*MAX {
            ans = min(ans, dp[t][c]);
        }
        println!("{}", ans);
    }
}