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"
#include<unordered_set>
#include<unordered_map>
#include<random>
using namespace std;
typedef long long ll;
const ll MOD = (ll)(1e9+7);
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define rep(i, n) for(int (i)=0; (i)<(int)(n); (i)++)
int dx[4]={ 1,0,-1,0 };
int dy[4]={ 0,1,0,-1 };

int N, M, S;
int U[100], V[100], A[100]; ll B[100];
int C[50]; ll D[50];

vector<pair<int, pair<int, ll>>> e[50];
ll cost[50][2501];

signed main(){
	cin >> N >> M >> S;
	rep(i, M){
		cin >> U[i] >> V[i] >> A[i] >> B[i];
		U[i]--; V[i]--;
		e[U[i]].pb({ V[i],{ A[i],B[i]} });
		e[V[i]].pb({ U[i],{ A[i],B[i]} });
	}
	rep(i, N) cin >> C[i] >> D[i];

	rep(i, N){
		rep(j, 2501) cost[i][j] = LLONG_MAX;
	}

	priority_queue<pair<ll, pair<int, int>>> pq;
	cost[0][min(S, 2500)] = 0;
	pq.push({ 0, {0, min(S, 2500)} });
	while(!pq.empty()){
		int n = pq.top().second.first;
		int s = pq.top().second.second;
		pq.pop();
		if(cost[n][min(s+C[n], 2500)] > cost[n][s] + D[n]){
			cost[n][min(s+C[n], 2500)] = cost[n][s] + D[n];
			pq.push({ -cost[n][min(s+C[n], 2500)] , {n, min(s+C[n], 2500)} });
		}
		rep(i, e[n].size()){
			if(e[n][i].second.first > s) continue;
			int m = e[n][i].first;
			int a = e[n][i].second.first;
			ll b = e[n][i].second.second;
			if(cost[m][s-a] > cost[n][s] + b){
				cost[m][s-a] = cost[n][s] + b;
				pq.push({ -cost[m][s-a] , {m, s-a} });
			}
		}
	}

	rep(i, N){
		if(i == 0) continue;
		ll ans = LLONG_MAX;
		rep(j, 2501){
			ans = min(ans, cost[i][j]);
		}
		cout << ans << endl;
	}
}