There is a simple undirected graph with N vertices and M edges. The vertices are numbered 1 through N, and the edges are numbered 1 through M. Edge i connects Vertex U_i and V_i. Also, Vertex i has two predetermined integers A_i and B_i. You will play the following game on this graph.

First, choose one vertex and stand on it, with W yen (the currency of Japan) in your pocket. Here, A_s \leq W must hold, where s is the vertex you choose. Then, perform the following two kinds of operations any number of times in any order:

* Choose one vertex v that is directly connected by an edge to the vertex you are standing on, and move to vertex v. Here, you need to have at least A_v yen in your pocket when you perform this move.
* Donate B_v yen to the vertex v you are standing on. Here, the amount of money in your pocket must not become less than 0 yen.



You win the game when you donate once to every vertex. Find the smallest initial amount of money W that enables you to win the game.

Constraints

* 1 \leq N \leq 10^5
* N-1 \leq M \leq 10^5
* 1 \leq A_i,B_i \leq 10^9
* 1 \leq U_i < V_i \leq N
* The given graph is connected and simple (there is at most one edge between any pair of vertices).

Input

Input is given from Standard Input in the following format:


N M
A_1 B_1
A_2 B_2
:
A_N B_N
U_1 V_1
U_2 V_2
:
U_M V_M


Output

Print the smallest initial amount of money W that enables you to win the game.

Examples

Input

4 5
3 1
1 2
4 1
6 2
1 2
2 3
2 4
1 4
3 4


Output

6


Input

5 8
6 4
15 13
15 19
15 1
20 7
1 3
1 4
1 5
2 3
2 4
2 5
3 5
4 5


Output

44


Input

9 10
131 2
98 79
242 32
231 38
382 82
224 22
140 88
209 70
164 64
6 8
1 6
1 4
1 3
4 7
4 9
3 7
3 9
5 9
2 5


Output

582
#include <bits/stdc++.h>
using namespace std;

int N, M;
#define MAXN 100000
long long B[MAXN], C[MAXN];
typedef tuple<pair<long long, int>, int, int> T;
vector<T> E;

int Rank[MAXN], Par[MAXN], Boss[MAXN];
vector<int> Kids[MAXN]; int Root;

typedef pair<long long, long long> P;

void init() {
  for(int v = 0; v < N; ++v) {
    Rank[v] = 0; Par[v] = Boss[v] = v;
  }
}
int find(int v) {
  if(Par[v] == v) return v;
  else return Par[v] = find(Par[v]);
}
void unite(int u, int v) {
  u = find(u); v = find(v);
  if(u == v) return;
  if(Rank[u] > Rank[v]) swap(u, v);
  Par[u] = v;
  if(Rank[u] == Rank[v]) ++Rank[v];
  if(make_pair(C[Boss[u]], Boss[u])
      > make_pair(C[Boss[v]], Boss[v])) {
    Kids[Boss[u]].push_back(Boss[v]);
    Boss[v] = Boss[u];
  } else {
    Kids[Boss[v]].push_back(Boss[u]);
  }
  Root = Boss[v];
}

P dp(int v) {
  long long b = B[v];
  vector<P> ps;
  for(int w : Kids[v]) {
    auto p = dp(w);
    ps.push_back(p); b += p.second;
  }
  long long q = C[v] + b;
  for(int i = 0; i < (int)ps.size(); ++i) {
    q = min(q, max(ps[i].first, C[v]) + b - ps[i].second);
  }
  return {q, b};
}

int main() {
  cin >> N >> M;
  for(int v = 0; v < N; ++v) {
    long long a, b; cin >> a >> b;
    B[v] = b; C[v] = max(a - b, 0LL);
  }
  for(int e = 0; e < M; ++e) {
    int u, v; cin >> u >> v; --u; --v;
    E.emplace_back(
      max(make_pair(C[u], u), make_pair(C[v], v)),
      u, v);
  }
  sort(E.begin(), E.end());
  init();
  for(int e = 0; e < M; ++e) {
    unite(get<1>(E[e]), get<2>(E[e]));
  }
  cout << dp(Root).first << endl;
  return 0;
}
