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
/*
[ ( ^ _ ^ ) ]
*/
// problem: atc/abc164_e
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define INF 1000000000

using edge = tuple<int, ll, ll>;

const int nax = 60;
const ll infll = LLONG_MAX/3;
vector<vector<edge>> g(nax);
vector<ll> c(nax);
vector<ll> d(nax);
int n, m;
ll s;

int main() {
    scanf("%d %d %lld", &n, &m, &s);
    for(int i=0; i<m; i++) {
        int u, v;
        ll a, b;
        scanf("%d %d %lld %lld", &u, &v, &a, &b);
        g[u].push_back({v, a, b});
        g[v].push_back({u, a, b});
    }
    for(int i=1; i<=n; i++) {
        scanf("%lld %lld", &c[i], &d[i]);
    }
    int mxn = 2500;
    vector<vector<ll>> dp(nax, vector<ll>(mxn, infll));
    s = min(mxn-1ll, s);
    dp[1][s] = 0;
    for(int i=0; i<=n; i++) {
        for(int v=1; v<=n; v++) {
            for(int p=0; p<mxn; p++) {
                if(p+c[v] < mxn) {
                    dp[v][p+c[v]] = min(dp[v][p+c[v]], dp[v][p] + d[v]);
                }
            }
            for(int p=0; p<mxn; p++) {
                if(dp[v][p] == infll) continue;
                for(auto [u, a, b]: g[v]) {
                    if(p - a >= 0) {
                        dp[u][p-a] = min(dp[u][p-a], dp[v][p] + b);
                    }
                }
            }
        }
    }
    for(int i=2; i<=n; i++) {
        ll rs = infll;
        for(int j=0; j<mxn; j++) {
            rs = min(rs, dp[i][j]);
        }
        printf("%lld\n", rs);
    }
}
