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>

const int N = 2e5 + 233;

int n, m, id[N];
long long A[N], B[N], C[N];
int vis[N], fa[N];
long long f[N], s[N];

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

std::vector<int> G[N], T[N];

void dfs(int x) {
  s[x] = B[x], f[x] = T[x].empty() ? std::max(A[x], B[x]) : 1e18;
  for (int y : T[x])
    dfs(y), s[x] += s[y];
  for (int y : T[x])
    f[x] = std::min(f[x], std::max(f[y], 1ll * C[x]) + s[x] - s[y]);
}

int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; ++i)
    scanf("%lld %lld", A + i, B + i), C[i] = A[i] - B[i], id[i] = i;
  for (int i = 1; i <= n; ++i)
    C[i] = C[i] < 0 ? 0 : C[i];
  for (int i = 1; i <= m; ++i) {
    int x, y; scanf("%d %d", &x, &y);
    G[x].push_back(y), G[y].push_back(x);
  }
  std::sort(id + 1, id + n + 1, [&](int x, int y)
    { return C[x] < C[y]; });
  for (int i = 1; i <= n; ++i) {
    int x = id[i]; vis[x] = 1;
    for (int y : G[x])
      if (vis[y] && find(y) != x) {
        T[x].push_back(find(y));
        fa[find(y)] = x;
      }
  }
  dfs(id[n]);
  std::cout << f[id[n]] << std::endl;
  return 0;
}

