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>
#define ll long long
using namespace std;
const int N = 55;
struct P{
    int v, a, b;
};
vector<P> G[N];
int c[N], d[N];
ll f[N][N * N];
int main() {
    int n, m, s;
    scanf("%d%d%d", &n, &m, &s), s = min(s, 2500);
    for(int i = 0, u, v, a, b; i < m; i++) {
        scanf("%d%d%d%d", &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("%d%d", &c[i], &d[i]);
    }
    for(int i = 0; i < N; i++) {
        for(int j = 0; j < N * N; j++) {
            f[i][j] = 1e18;
        }
    }
    f[1][s] = 0;
    typedef tuple<ll, int, int> T;
    priority_queue<T, vector<T>, greater<T>> qu;
    qu.push(T(0, 1, s));
    while(!qu.empty()) {
        ll t = get<0>(qu.top());
        int u = get<1>(qu.top()), w = get<2>(qu.top());
        qu.pop();
        if(f[u][w] > t) continue;
        for(int i = 0; i < G[u].size(); i++) {
            int v = G[u][i].v, a = G[u][i].a, b = G[u][i].b;
            if(w >= a && f[v][w - a] > t + b) {
                f[v][w - a] = t + b;
                qu.push(T(t + b, v, w - a));
            }
        }
        if(f[u][min(w + c[u], 2500)] > t + d[u]) {
            f[u][min(w + c[u], 2500)] = t + d[u];
            qu.push(T(t + d[u], u, min(w + c[u], 2500)));
        }
    }
    for(int i = 2; i <= n; i++) {
        ll ans = 1e18;
        for(int j = 0; j <= 2500; j++) {
            ans = min(ans, f[i][j]);
        }
        printf("%lld\n", ans);
    }
    return 0;
}

