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 <iostream>
#include <vector>
#include <cstring>

#define LL long long
using namespace std;
#define MAXN 60
#define MAXA 60

int N, M, S;

struct edge
{
    int v, a, b;
};
vector<edge> G[MAXN];
int C[MAXN], D[MAXN];
LL dp[MAXN][MAXA * MAXN];
LL ans;

inline void add(int u, int v, int a, int b)
{
    G[u].push_back({v, a, b});
    G[v].push_back({u, a, b});
}
int main()
{
    scanf("%d%d%d", &N, &M, &S);
    int u, v, a, b;
    for (int i = 1; i <= M; i++)
    {
        scanf("%d%d%d%d", &u, &v, &a, &b);
        add(u, v, a, b);
    }
    for (int i = 1; i <= N; i++)
    {
        scanf("%d%d", &C[i], &D[i]);
    }

    memset(dp, 0x3f, sizeof(dp));
    if (S > 2500)
        S = 2500;
    dp[1][S] = 0;
    for (int t = 0; t <= N; t++)
    {
        for (int i = 1; i <= N; i++)
        {
            for (int k = 0; k < G[i].size(); k++)
            {
                for (int j = 0; j <= 2500; j++)
                {
                    if (j + G[i][k].a <= 2500)
                        dp[i][j] = min(dp[i][j], dp[G[i][k].v][j + G[i][k].a] + G[i][k].b);
                }
            }

            for (int j = 0; j <= 2500; j++)
            {
                if (j >= C[i])
                    dp[i][j] = min(dp[i][j], dp[i][j - C[i]] + D[i]);
            }
        }
    }
    for (int i = 2; i <= N; i++)
    {
        ans = 1e18;
        for(int j=0; j<=2500; j++) {
            if(ans>dp[i][j]) {
                ans = dp[i][j];
            }
        }
        cout << ans << endl;
    }

    return 0;
}