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;

const int N = 55;
const int NN = N * N * 5;
vector<tuple<int, int, int>> g[N];
int c[N], d[N];
long long dp[N][NN];

priority_queue<tuple<long long, int, int>> q;

int add(int a, int b) {
  return min(a + b, NN - 1);
}

long long add2(long long a, long long b) {
  if (LLONG_MAX - a < b) {
    return LLONG_MAX;
  } else {
    return a + b;
  }
}

void push(long long d, int i, int j) {
  if (dp[i][j] > d) {
    dp[i][j] = d;
    q.emplace(-d, i, j);
  }
}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);

  for (int i = 0; i < N; ++i) {
    for (int j = 0; j < NN; ++j) {
      dp[i][j] = LLONG_MAX;
    }
  }

  int n, m, s;
  cin >> n >> m >> s;
  s = add(0, s);
  while (m--) {
    int u, v, a, b;
    cin >> u >> v >> a >> b;
    --u, --v;
    g[u].emplace_back(v, a, b);
    g[v].emplace_back(u, a, b);
  }
  for (int i = 0; i < n; ++i) {
    cin >> c[i] >> d[i];
  }
  push(0, 0, s);
  while (!q.empty()) {
    long long dd;
    int i, j;
    tie(dd, i, j) = q.top();
    q.pop();
    dd = -dd;
    if (dp[i][j] < dd) {
      continue;
    }
    push(add2(dd, d[i]), i, add(j, c[i]));
    for (auto [v, a, b] : g[i]) {
      if (j < a) {
        continue;
      }
      push(add2(dd, b), v, j - a);
    }
  }
  for (int i = 1; i < n; ++i) {
    cout << *min_element(dp[i], dp[i] + NN) << '\n';
  }

  return 0;
}
