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;

const int N = 1e5 + 5;
const int M = 2e5 + 5;

int fir[N] , ne[M] , to[M] , cnt , x , y , A[N] , B[N] , fa[N] , id[N] , n , m;
long long S[N] , More[N];

void add(int x , int y) {
  ne[++ cnt] = fir[x];
  fir[x] = cnt;
  to[cnt] = y;
}

void link(int x , int y) {
  add(x , y);
  add(y , x);
}

#define Foreachson(i , x) for(int i = fir[x];i;i = ne[i])

int getf(int x) {
  return (fa[x] == x) ? x : (fa[x] = getf(fa[x]));
}

bool cmp(int x , int y) {
  return A[x] < A[y];
}

bool vis[N];

int main() {
  scanf("%d%d" , &n , &m);
  int it = 0;
  for(int i = 1;i <= n;++ i) {
    scanf("%d%d" , &A[i] , &B[i]);
    More[i] = A[i] = max(0 , A[i] - B[i]);
    fa[i] = id[i] = i;
    S[i] = B[i];
  }
  for(int i = 1;i <= m;++ i) {
    scanf("%d%d" , &x , &y);
    link(x , y);
  }
  sort(id + 1 , id + n + 1 , cmp);
  for(int i = 1;i <= n;++ i) {
    int x = id[i];
    vis[x] = 1;
    Foreachson(j , x) {
      int V = to[j];
      if(!vis[V])
	continue;
      int a = getf(x) , b = getf(V);
      if(a == b)
	continue;
      fa[b] = a; S[a] += S[b];
      More[a] = min(More[a] , More[b] + max(0LL , A[x] - More[b] - S[b]));
    }
  }
  int now = getf(1);
  cout << S[now] + More[now] << endl;
}
