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;
typedef long long ll;

struct edge {
    ll v, a, b; // v - vertex, a - edge cost, b - time
};

struct entry {
    ll b, v, c; // b - time, v - vertex, c - coins
};

bool operator>(const entry &l, const entry &r) {
    return l.b > r.b;
}

const ll N = 52, A = 2510;
vector<edge> g[N];
pair<ll, ll> cd[N];
ll dp[N][A + 1];

int main() {
    ll n, m, s;
    cin >> n >> m >> s;
    for (ll i = 1; i <= m; i++) {
        ll u, v, a, b;
        cin >> u >> v >> a >> b;
        g[u].push_back({v, a, b});
        g[v].push_back({u, a, b});
    }
    for (ll i = 1; i <= n; i++) cin >> cd[i].first >> cd[i].second;
    priority_queue<entry, vector<entry>, greater<entry>> q;
    q.push({0, 1, min(s, A)});
    while (!q.empty()) {
        auto entry = q.top();
        q.pop();
        if (dp[entry.v][entry.c]) continue;
        dp[entry.v][entry.c] = entry.b;
        for (auto edge : g[entry.v]) {
            if (entry.c - edge.a >= 0) {
                q.push({entry.b + edge.b, edge.v, min(entry.c - edge.a, A)});
            }
        }
        q.push({entry.b + cd[entry.v].second, entry.v, min(entry.c + cd[entry.v].first, A)});
    }
    for (int i = 2; i <= n; i++) {
        ll ans = 1e16;
        for (int j = 0; j <= A; j++) if (dp[i][j]) ans = min(ans, dp[i][j]);
        cout << ans << "\n";
    }
}
