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;

typedef long long ll;
const int N = 100233;
int n, m, fa[N], p[N], visit[N];
ll a[N], b[N], f[N], s[N];
vector<int> G[N], E[N];
int F(int x) {while (x != fa[x]) x = fa[x] = fa[fa[x]]; return x;}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%lld%lld", &a[i], &b[i]);
  }
  for (int i = 1, x, y; i <= m; i++) {
    scanf("%d%d", &x, &y);
    G[x].push_back(y);
    G[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    fa[i] = p[i] = i;
  }
  sort(p + 1, p + 1 + n, [&](const int &x, const int &y){return a[x] - b[x] < a[y] - b[y];});
  for (int i = 1; i <= n; i++) {
    int x = p[i], z;
    visit[x] = 1;
    s[x] = b[x];
    for (auto y : G[x]) if (visit[y]) {
      if ((z = F(y)) == x) continue;
      fa[z] = x;
      s[x] += s[z];
      E[x].push_back(z);
    }
  }
  for (int i = 1; i <= n; i++) {
    int x = p[i];
    if (E[x].empty()) f[x] = max(a[x], b[x]);
    else f[x] = 1ll << 60;
    for (auto y : E[x]) {
      f[x] = min(f[x], max(f[y], a[x] - b[x]) + s[x] - s[y]);
    }
  }
  printf("%lld\n", f[p[n]]);
}
