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 proconio::input;
use proconio::marker::Usize1;
use std::cmp::{Reverse, min};
use std::collections::BinaryHeap;

struct Path {
    node: usize,
    cost: usize,
    time: u64,
}

impl Clone for Path {
    fn clone(&self) -> Path {
        Path{node: (&self).node, cost: (&self).cost, time: (&self).time}
    }
}

fn main() {
    input!{
        n: usize,
        m: usize,
        s: usize,
        edges: [(Usize1, Usize1, usize, u64); m],
        ex_infos: [(usize, u64);n],
    }
    
    let mut graph: Vec<Vec<Path>> = vec![vec![]; n];
    let mut all_cost = 0;
    for (u, v, cost, time) in edges {
        graph[u].push(Path{node: v, cost, time});
        graph[v].push(Path{node: u, cost, time});
        all_cost += cost;
    }

    solve(graph, ex_infos, s, all_cost);
}

fn solve(graph: Vec<Vec<Path>>, exchange: Vec<(usize, u64)>, s: usize, max_cost: usize) -> () {
    let max_time = std::u64::MAX;
    let s = min(max_cost, s);
    let n = graph.len();
    let mut dp = vec![vec![max_time; max_cost + 1]; n];
    let mut is_fixed = vec![vec![false; max_cost + 1]; n];

    // heap(所要時間, 所持銀貨, 都市)
    let mut r_heap: BinaryHeap<(Reverse<u64>, usize, usize)> = BinaryHeap::new();
    
    r_heap.push((Reverse(0), s, 0));
    while !r_heap.is_empty() {
        let (Reverse(time), poss, node) = r_heap.pop().unwrap();
        if is_fixed[node][poss] {continue};
        
        // 確定
        for i in 0..poss + 1 {
            if dp[node][i] > time {
                dp[node][i] = time; 
                is_fixed[node][i] = true;
            }
        }

        // 両替
        let (rate, ex_time) = exchange[node];
        let (exd_poss, exd_time) = (min(poss + rate, max_cost), time + ex_time);
        //if dp[node][exd_poss] > exd_time {
        if !is_fixed[node][exd_poss] {
            r_heap.push((Reverse(exd_time), exd_poss, node));
        }
        
        // 移動
        for next in &graph[node] {
            let next_time = next.time + dp[node][poss];
            if next.cost > poss 
                || is_fixed[next.node][poss - next.cost] {continue;} 
            //|| dp[next.node][poss - next.cost] < next_time {continue;};
            r_heap.push((Reverse(next_time), poss - next.cost, next.node));

        }
    }

    let ans: Vec<_> = dp.iter().map(|v| v.iter().min().unwrap()).collect();
    for a in &ans[1..] {
        println!("{}", a);
    }
}