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.

Examples

Input

3 2 1
1 2 1 2
1 3 2 4
1 11
1 2
2 5


Output

2
14


Input

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


Output

5
5
7


Input

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


Output

1
9003
14606
16510
16576


Input

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


Output

1
3
5


Input

2 1 0
1 2 1 1
1 1000000000
1 1


Output

1000000001
#include <bits/stdc++.h>
using namespace std;

const int LIM = 2500;
const long long INF = (long long) 1e18;

int C[55], D[55];
int done[55][2505];
long long SP[55][2505];

vector< vector<int> > X[55];
priority_queue< vector<long long>, vector< vector<long long> >, greater< vector<long long> > > Q;

void relax(int ux, int uy, int vx, int vy, int w)
{
    if(SP[vx][vy] > SP[ux][uy] + w)
    {
        SP[vx][vy] = SP[ux][uy] + w;
        Q.push({SP[vx][vy], vx, vy});
    }
}

int main() {
    int n, m, s;
    cin >> n >> m >> s;

    while(m--)
    {
        int u, v, a, b;
        cin >> u >> v >> a >> b;

        X[u].push_back({v, a, b});
        X[v].push_back({u, a, b});
    }

    for (int i=1; i<=n; i++)
        cin >> C[i] >> D[i];

    for (int i=1; i<=n; i++)
        for (int j=0; j<=LIM; j++)
            SP[i][j] = INF;
    
    s = min(s, LIM);
    SP[1][s] = 0;
    Q.push({SP[1][s], 1, s});

    while(!Q.empty()) {
        auto q = Q.top();
        Q.pop();

        int indx = q[1], coins = q[2];

        if(done[indx][coins])
            continue;
        
        done[indx][coins] = 1;

        for (int i=1; coins+i<=LIM && i<=C[indx]; i++)
            if (!done[indx][coins+i])
                relax(indx, coins, indx, coins + i, D[indx]);
        
        for (auto x : X[indx])
            if (x[1] <= coins && !done[ x[0] ][ coins - x[1] ])
                relax(indx, coins, x[0], coins - x[1], x[2]);
    }

    for (int i=2; i<=n; i++)
        cout << *min_element(SP[i], SP[i]+LIM+1) << "\n";

    return 0;
}
