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 st first
#define nd second

using ll = long long ;
using namespace std;
using pii = pair<int,int>;
const int maxn = 5000;
const ll INF = 1e18 + 7;
ll dp[55][5050], ans[55];
struct path{
	int st;
	int nd;
	ll val;
	bool operator < (const path &o) const{
		return val > o.val;
	}
};


std::vector<path> gr[60];
pii ex[60];

int main(){
	int n, m, sil, x, y;
	ll z;
	scanf("%d %d %d", &n, &m, &sil);
	for(int i = 1; i <= m; i++){
		int u, v, a;
		ll b;
		scanf("%d %d %d %lld", &u, &v, &a, &b);
		gr[u].push_back({v, a, b});
		gr[v].push_back({u, a, b});
	}
	for(int i = 1; i <= n; i++){
		ans[i] = INF;
		for(int j = 0; j <= 5000; j++){
			dp[i][j] = INF;
		}
	}
	for(int i = 1; i <= n; i ++){
		scanf("%d %d", &ex[i].st, &ex[i].nd);
	}
	priority_queue<path> pq;
	pq.push({1,min(sil, maxn), 0});
	int cnt = 0;
	while(!pq.empty()){
		path p = pq.top();
		pq.pop();
		x  = p.st, y = p.nd, z = p.val;
		if(dp[x][y] != INF){
			continue;
		}
		//printf("at %d %d %lld\n", x, y, z);
		dp[x][y] = z;
		if(ans[x] == INF) ans[x] = z, cnt++;
		if(cnt == n) break;
		int ma = min(y + ex[x].st, 5000);
		if(dp[x][ma] == INF) pq.push({x, ma, z + ex[x].nd});
		for(auto it: gr[x]){
			if(it.nd > y) continue;
			if(dp[it.st][y - it.nd] == INF){
				pq.push({it.st, y - it.nd, z + it.val});
			}
		}
	}
	for(int i = 2; i <= n; i++){
		printf("%lld\n", ans[i]);
	}

}