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 ll long long
#define fornum(A,B,C) for(A=B;A<C;++A)
#define pii pair<int,int>
#define pll pair<ll,ll>

using namespace std;

/////////////////////////////////////////////////////

ll N,M,S;
ll C[111],D[111];
vector<pair<ll,pll>> uvv[1111];

ll i, j, k;

void query(ll aa){
    ll i,j,k;
    priority_queue<pair<ll,pll>> q;
    vector<vector<bool>> mk(N + 5, vector<bool>(5555));
    q.push({0, {1, S > 5000 ? 5000 : S}});
    while(q.size()){
        ll d = q.top().first;
        pll qa = q.top().second;
        q.pop();
        ll a = qa.first;
        ll c = qa.second;
        if(a==aa){
            printf("%lld\n", -d);
            return;
        }
        if(mk[a][c])
            continue;
        mk[a][c] = true;
        if(c<5000){
            q.push({d - D[a], {a, min(5000ll, c + C[a])}});
        }
        fornum(i,0,uvv[a].size()){
            auto bbb = uvv[a][i];
            ll bb = bbb.first;
            ll cc = bbb.second.first;
            ll dd = bbb.second.second;
            if(c<cc)
                continue;
            if(mk[bb][c-cc])
                continue;
            q.push({d - dd, {bb, c - cc}});
        }
    }
}

int main(){
    scanf("%lld%lld%lld", &N, &M, &S);
    fornum(i,0,M){
        ll u, v, a, b;
        scanf("%lld%lld%lld%lld", &u, &v, &a, &b);
        uvv[u].push_back({v, {a, b}});
        uvv[v].push_back({u, {a, b}});
    }
    fornum(i,0,N){
        scanf("%lld%lld", &C[i+1], &D[i+1]);
    }

    fornum(i,1,N){
        query(i + 1);
    }
    return 0;
}