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;
typedef long long ll;
const int maxn=55;
const int maxm=2466;
const ll inf=0x3f3f3f3f3f3f3f3f;//long long最大值，8个3f
#define tp tuple<ll,ll,ll>

vector<tp> G[maxn];//存每个点所能到达的点
ll c[maxn],d[maxn];//此点换钱的钱数和花费时间
ll dp[maxn][maxm];

int main(){
	int n,m,s;
	scanf("%d%d%d",&n,&m,&s);
	s=min(2450,s);//防止越界，取2450（最大容量）以下。
	for(int i=1;i<=m;i++){
		ll u,v,a,b;
		scanf("%lld%lld%lld%lld",&u,&v,&a,&b);
		G[u].push_back(tp(v,a,b)); //点 钱 时间 
		G[v].push_back(tp(u,a,b));
	}
	for(int i=1;i<=n;i++)
		scanf("%d%d",&c[i],&d[i]);
	fill(dp[0],dp[0]+54*2465,inf);//全部初始化为inf的骚操作
	dp[1][s]=0;//初始化起点
	//优先队列，按照tuple三个值（从前往后）升序
	priority_queue<tp,vector<tp>,greater<tp>> pq;
	pq.push(tp(0,1,s));
	
	while(!pq.empty()){
		tp cur;
		cur=pq.top();
		pq.pop();
		ll t=get<0>(cur),u=get<1>(cur),w=get<2>(cur);//取tuple值
		if(dp[u][w]>t) continue; //可有可无的判重
		for(auto i:G[u]){//遍历vector
			ll v=get<0>(i),toa=get<1>(i),tob=get<2>(i);
			if(dp[v][w-toa]>t+tob&&w>=toa){//如果有足够钱可达且这样走比原有dp更快
				dp[v][w-toa]=t+tob;
				pq.push(tp(t+tob,v,w-toa));//存入状态
			}
		}
		if(dp[u][min(2450LL,w+c[u])]>t+d[u]){//如果充1次钱后比原有dp更快
			dp[u][min(2450LL,w+c[u])]=t+d[u];
			pq.push(tp(t+d[u],u,min(2450LL,w+c[u])));//存入状态
		}
		//这里只充一次钱，因为接下来会有下一次充钱的状态到来~
	}
	for(int i=2;i<=n;i++){
		ll ans=inf;
		for(int j=0;j<=2450;j++)
			ans=min(ans,dp[i][j]);//取到达时剩下的不同钱中最快的
		printf("%lld\n",ans);
	}
}
