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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define int long long
#define rep(i, n) for (int i = 0; i < (n); i++)
#define reps(i, n) for (int i = 1; i <= (n); i++)
#define all(x) begin(x), end(x)
#define uniq(x) (x).erase(unique(all(x)), end(x))
#define bit(n) (1LL << (n))
#define cdiv(a, b) (((a) - 1) / (b) + 1)
#define dump(x) cerr << #x " = " << (x) << endl
using vint = vector<int>;
using vvint = vector<vint>;
using pint = pair<int, int>;
using vpint = vector<pint>;
template<typename T> using priority_queue_rev = priority_queue<T, vector<T>, greater<T>>;
constexpr long double PI = 3.1415926535897932384626433832795028L;
constexpr int DY[8] = {0, 1, 0, -1, 1, 1, -1, -1};
constexpr int DX[8] = {1, 0, -1, 0, 1, -1, -1, 1};
int gcd(int a, int b) {
    while (b) { swap(a %= b, b); }
    return a;
}
int lcm(int a, int b) { return a / gcd(a, b) * b; }
template<typename T> void fin(T mes) {
    cout << mes << endl;
    exit(0);
}
template<typename T, typename U> bool chmax(T &a, const U &b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
template<typename T, typename U> bool chmin(T &a, const U &b) {
    if (b < a) {
        a = b;
        return true;
    }
    return false;
}
template<typename T, typename U> ostream &operator<<(ostream &os, const pair<T, U> &rhs) {
    os << "(" << rhs.first << ", " << rhs.second << ")";
    return os;
}
template<typename T> ostream &operator<<(ostream &os, const vector<T> &rhs) {
    os << "{";
    for (auto itr = rhs.begin(); itr != rhs.end(); itr++) {
        os << *itr << (next(itr) != rhs.end() ? ", " : "");
    }
    os << "}";
    return os;
}
struct setup {
    static constexpr int PREC = 20;
    setup() {
        cout << fixed << setprecision(PREC);
        cerr << fixed << setprecision(PREC);
    };
} setup;

int N, M, S;
int U[110], V[110], A[110], B[110];
int C[55], D[55];
int dp[55][5500];
signed main() {
    cin >> N >> M >> S;
    reps(i, M) { cin >> U[i] >> V[i] >> A[i] >> B[i]; }
    reps(i, N) { cin >> C[i] >> D[i]; }
    rep(i, 55)rep(j, 5500) { dp[i][j] = LLONG_MAX; }
    dp[1][min(S, 5499LL)] = 0;
    reps(i, 2 * M) {
        reps(j, N)rep(k, 5500) { if (dp[j][k] < LLONG_MAX) { chmin(dp[j][min(k + C[j], 5499LL)], dp[j][k] + D[j]); }}
        reps(j, M) {
            for (int k = A[j]; k < 5500; k++) {
                if (dp[V[j]][k] < LLONG_MAX) { chmin(dp[U[j]][k - A[j]], dp[V[j]][k] + B[j]); }
                if (dp[U[j]][k] < LLONG_MAX) { chmin(dp[V[j]][k - A[j]], dp[U[j]][k] + B[j]); }
            }
        }
    }
    for (int i = 2; i <= N; i++) {
        int ans = LLONG_MAX;
        rep(j, 5500) { chmin(ans, dp[i][j]); }
        cout << ans << endl;
    }
}