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 ll = long long;
const int maxn = 52, maxm = 103, maxa = 52;

int n, m;
ll starts, c[maxn], d[maxn], dp[maxn][maxa * maxm];
std::vector<std::tuple<int, ll, ll>> E[maxn];

std::priority_queue<std::tuple<ll, int, int>> q;

void update(int i, int c, ll w) {
    if(c < 0 || c >= maxa * maxm || dp[i][c] <= w)
        return;
    dp[i][c] = w;
    q.emplace(-w, i, c);
}

void solve() {
    update(1, std::min(starts, maxa * (ll)maxm - 1), 0);
    while(!q.empty()) {
        ll w; int i, mc;
        std::tie(w, i, mc) = q.top();
        q.pop();
        w = -w;
        if(w != dp[i][mc])
            continue;
        // fprintf(stderr, "%d %d %lld\n", i, mc, w);
        update(i, mc + c[i], w + d[i]);
        for(auto &&[v, cost, minute] : E[i])
            update(v, mc - cost, w + minute);
    }
}

int main() {
    scanf("%d%d%lld", &n, &m, &starts);
    for(int i = 0; i < m; ++i) {
        int u, v;
        ll a, b;
        scanf("%d%d%lld%lld", &u, &v, &a, &b);
        E[u].emplace_back(v, a, b);
        E[v].emplace_back(u, a, b);
    }
    for(int i = 1; i <= n; ++i) {
        scanf("%lld%lld", c + i, d + i);
    }
    memset(dp, 0x7f, sizeof(dp));
    solve();
    for(int i = 2; i <= n; ++i)
        printf("%lld\n", *std::min_element(dp[i], dp[i] + maxa * maxm));
    return 0;
}