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(clippy::needless_range_loop)]
#![allow(unused_macros)]
#![allow(dead_code)]

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        #[allow(unused_mut)]
        let mut s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };

    ($iter:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

use std::cmp::*;
use std::collections::binary_heap::BinaryHeap;
#[derive(Copy, Clone, Eq, PartialEq)]
struct GraphEdge {
    node_to: usize,
    cost: isize,
}
impl Ord for GraphEdge {
    fn cmp(&self, other: &GraphEdge) -> Ordering {
        other.cost.cmp(&self.cost)
    }
}
impl PartialOrd for GraphEdge {
    fn partial_cmp(&self, other: &GraphEdge) -> Option<Ordering> {
        Some(self.cmp(other))
    }
}
impl GraphEdge {
    fn new(node_to: usize, cost: isize) -> Self {
        Self { node_to, cost }
    }
}
struct Graph {
    pub edges: Vec<Vec<GraphEdge>>,
    pub nodes: usize,
}
impl Graph {
    fn new(nodes: usize) -> Self {
        Self {
            edges: vec![vec![]; nodes],
            nodes,
        }
    }
    fn edge_undirected(&mut self, node_a: usize, node_b: usize) {
        self.edges[node_a].push(GraphEdge::new(node_b, 1));
        self.edges[node_b].push(GraphEdge::new(node_a, 1));
    }
    fn edge_directed(&mut self, node_from: usize, node_to: usize) {
        self.edges[node_from].push(GraphEdge::new(node_to, 1));
    }
    fn edge_undirected_costed(&mut self, node_a: usize, node_b: usize, cost: isize) {
        self.edges[node_a].push(GraphEdge::new(node_b, cost));
        self.edges[node_b].push(GraphEdge::new(node_a, cost));
    }
    fn edge_directed_costed(&mut self, node_from: usize, node_to: usize, cost: isize) {
        self.edges[node_from].push(GraphEdge::new(node_to, cost));
    }
    fn dijkstra(&self, node_from: usize) -> Vec<isize> {
        let mut costs = vec![std::isize::MAX; self.nodes];
        let mut heap: BinaryHeap<GraphEdge> = BinaryHeap::new();
        heap.push(GraphEdge::new(node_from, 0));
        while let Some(GraphEdge { cost, node_to }) = heap.pop() {
            if cost >= costs[node_to] {
                continue;
            }
            costs[node_to] = cost;
            for edge in &self.edges[node_to] {
                let next = GraphEdge::new(edge.node_to, cost + edge.cost);
                heap.push(next);
            }
        }
        costs
    }
}

const A_MAX: usize = 2500;

fn main() {
    input! {
        n: usize,
        m: usize,
        s: usize,
        edges: [(usize1, usize1, usize, usize); m],
        exchs: [(usize, usize); n],
    }
    let s = std::cmp::min(A_MAX, s);
    let mut graph = Graph::new(n * (A_MAX + 1));
    for (left, right, fare, time) in &edges {
        for i in *fare..=A_MAX {
            graph.edge_directed_costed(
                *left * (A_MAX + 1) + i,
                *right * (A_MAX + 1) + i - fare,
                *time as isize,
            );
            graph.edge_directed_costed(
                *right * (A_MAX + 1) + i,
                *left * (A_MAX + 1) + i - fare,
                *time as isize,
            );
        }
    }
    for (index, (coins, time)) in exchs.iter().enumerate() {
        for i in 0..=(A_MAX - *coins) {
            graph.edge_directed_costed(
                index * (A_MAX + 1) + i,
                index * (A_MAX + 1) + i + *coins,
                *time as isize,
            );
        }
    }
    let costs = graph.dijkstra(s);
    let costs_min: Vec<_> = (1..n)
        .map(|city| {
            costs[(city * (A_MAX + 1))..((city + 1) * (A_MAX + 1))]
                .iter()
                .min()
                .unwrap()
        })
        .copied()
        .collect();
    
    for ans in &costs_min {
        println!("{}", ans);
    }
}