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;
using ll = long long;
using vl = vector<ll>;
using pll = pair<ll,ll>;
using plll = pair<ll,pll>;
using vll = vector<pll>;
using vvll = vector<vll>;

#define ff first
#define ss second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define dmin(a,b) a = min(a,b)

const ll INF = 1e18;

int main() {
    ll n, m, s; cin >> n >> m >> s;
    vvll edges(n, vll(n, pll(INF, INF)));
    vl ans(n, INF);
    rep(i,m) {
        ll u, v, a, b; cin >> u >> v >> a >> b;
        u--; v--;
        edges[u][v] = edges[v][u] = pll(a, b);
    }
    rep(i,n) {
        ll a, b; cin >> a >> b;
        edges[i][i] = pll(-a, b);
    }
    priority_queue<plll> pq;
    map<pll,ll> used;
    pq.push(plll(0, pll(0, s)));
    while (!pq.empty()) {
        ll time = -pq.top().ff;
        ll node = pq.top().ss.ff;
        ll cnt  = pq.top().ss.ss; pq.pop();
        if (cnt < 0) continue;
        dmin(cnt, 2500LL);
        if (used.count(pll(node,cnt))) continue;
        used[pll(node,cnt)] = 1;
        dmin(ans[node], time);
        rep(to,n) {
            ll a = edges[node][to].ff, b = edges[node][to].ss;
            if (a == INF) continue;
            pq.push(plll(-time-b, pll(to, cnt-a)));
        }

        bool finished = true;
        rep(i,n) if (ans[i] == INF) finished = false;
        if (finished) break;
    }
    rep(i,n-1) cout << ans[i+1] << endl;
    return 0;
}
