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;

using ll = long long;
int main() {
   ios_base::sync_with_stdio(false);
   int N, M;
   cin >> N >> M;
   vector<int> A(N), B(N);
   vector<int> par(N);
   vector<ll> sum(N);
   vector<ll> ans(N);
   for (int i = 0; i < N; ++i) {
      cin >> A[i] >> B[i];
      par[i] = i, sum[i] = B[i], ans[i] = max(A[i], B[i]);
   }
   vector<vector<int>> adj(N);
   for (int i = 0; i < M; ++i) {
      int x, y;
      cin >> x >> y;
      --x, --y;
      adj[x].emplace_back(y);
      adj[y].emplace_back(x);
   }
   vector<int> ord(N);
   iota(ord.begin(), ord.end(), 0);
   sort(ord.begin(), ord.end(), [&](int x, int y) {
      return A[x] - B[x] < A[y] - B[y];
   });
   function<int(int)> getPar = [&](int v) {
      if (v != par[v]) {
         par[v] = getPar(par[v]);
      }
      return par[v];
   };
   vector<bool> visited(N);
   for (int v : ord) {
      vector<int> nxts;
      for (int u : adj[v]) {
         if (visited[u]) nxts.emplace_back(getPar(u));
      }
      sort(nxts.begin(), nxts.end());
      nxts.resize(unique(nxts.begin(), nxts.end()) - nxts.begin());
      for (int u : nxts) {
         par[u] = v;
         sum[v] += sum[u];
      }
      if (nxts.size()) {
         ans[v] = 1e18;
         for (int u : nxts) {
            ans[v] = min(ans[v], max(ll(A[v] - B[v]), ans[u]) + sum[v] - sum[u]);
         }
      }
      visited[v] = true;
   }
   cout << ans[ord.back()] << "\n";
   return 0;
}
