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 JavaScript (Node.js 12.16.1) (judge verdict: Accepted)

"use strict";
var input=require("fs").readFileSync("/dev/stdin","utf8");
var cin=input.split(/ |\n/),cid=0;
function next(a){return a?cin[cid++]:+cin[cid++];}
function nexts(n,a){return a?cin.slice(cid,cid+=n):cin.slice(cid,cid+=n).map(a=>+a);}
function nextm(h,w,a){var r=[],i=0;if(a)for(;i<h;i++)r.push(cin.slice(cid,cid+=w));else for(;i<h;i++)r.push(cin.slice(cid,cid+=w).map(a=>+a));return r;}
function xArray(v){var a=arguments,l=a.length,r="Array(a["+--l+"]).fill().map(x=>{return "+v+";})";while(--l)r="Array(a["+l+"]).fill().map(x=>"+r+")";return eval(r);}

class dijkstra{
  constructor(n){
    this.graph = Array(n).fill().map(a=>[]);
  }
  add(u,v,cost){
    this.graph[u].push(v,cost);
  }
  add2(u,v,cost){
    this.graph[u].push(v,cost);
    this.graph[v].push(u,cost);
  }
  solve(start){
    const INF = 1e18;
    const g = this.graph;
    var pq = new priorityQueue();
    pq.push([0,start]);
    var dp = Array(g.length).fill(INF);
    dp[start] = 0;
    while(pq.size){
      var [t,v] = pq.pop();
      if(dp[v] < t)continue;
      for(var i = 0; i < g[v].length; i+=2){
        var V = g[v][i], c = g[v][i+1] + t;
        if(dp[V] <= c)continue;
        dp[V] = c;
        pq.push([c,V]);
      }
    }
    return dp;
  }
}

class priorityQueue{
  constructor(){
    this.a=[];
  }
  push(x){
    const a = this.a;
    for(var i = a.length, j; i; i = j){
      j = i - 1 >> 1;
      if(x[0] >= a[j][0])break;
      a[i] = a[j];
    }
    a[i] = x;
  }
  pop(){
    const a = this.a;
    var r = a[0];
    var x = a.pop();
    var k = a.length >> 1;
    for(var i = 0, j; i < k; i = j){
      j = (i << 1) + 1;
      if(a[j+1] && a[j][0] > a[j+1][0])j++;
      if(a[j][0] >= x[0])break;
      a[i] = a[j];
      i = j;
    }
    if(a.length)a[i] = x;
    return r;
  }
  get size(){
    return this.a.length;
  }
  get top(){
    return this.a[0];
  }
}

const mx = 2500;

var myOut = main();
if(myOut !== undefined)console.log(myOut);

function main(){
  var [n,m,s] = nexts(3);
  var g = new dijkstra(mx * n);
  for(var i = 0; i < m; i++){
    var [u,v,a,b] = nexts(4);
    u--,v--;
    u *= mx, v *= mx;
    for(var j = mx; j-- > a;){
      g.add(u+j,v+j-a,b);
      g.add(v+j,u+j-a,b);
    }
  }
  for(var i = 0; i < n; i++){
    var [c,d] = nexts(2);
    var v = i * mx;
    for(var j = 0; j < mx-1; j++)g.add(v+j,v+Math.min(mx-1,j+c),d);
  }
  var dp = g.solve(Math.min(mx-1,s));
  var ans = xArray(1e18,n);
  for(var i = 1; i < n; i++){
    var v = i*mx;
    for(var j = 0; j < mx; j++){
      ans[i] = Math.min(ans[i],dp[v+j]);
    }
  }
  return ans.slice(1).join("\n");
}