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>

void optimize_cpp_stdio() {
    std::ios::sync_with_stdio(false);
    std::cout.tie(nullptr);
    std::cin.tie(nullptr);
}

int main() {
    optimize_cpp_stdio();
    int n, m;
    std::cin >> n >> m;
    std::vector<int> a(n), b(n), c(n);
    for (int i = 0; i < n; i++) {
        std::cin >> a[i] >> b[i];
        c[i] = std::max(a[i] - b[i], 0);
    }
    std::vector<std::vector<int>> G(n);
    for (int i = 0, u, v; i < m; i++) {
        std::cin >> u >> v;
        G[--u].push_back(--v);
        G[v].push_back(u);
    }
    std::vector<int> parent(n);
    std::iota(parent.begin(), parent.end(), 0);
    std::function<int(int)> find = [&](int x) {
        return x == parent[x] ? x : parent[x] = find(parent[x]);
    };
    std::vector<int> sorted(n);
    std::iota(sorted.begin(), sorted.end(), 0);
    std::sort(sorted.begin(), sorted.end(), [&](int lhs, int rhs) {
        return c[lhs] < c[rhs];
    });
    std::vector<bool> vis(n);
    std::vector<int64_t> dp(c.begin(), c.end());
    std::vector<int64_t> cost(b.begin(), b.end());
    for (int u : sorted) {
        vis[u] = true;
        for (int v : G[u]) {
            v = find(v);
            if (not vis[v] || u == v) continue;
            parent[v] = u;
            cost[u] += cost[v];
            dp[u] = std::min(dp[u], std::max(dp[v], c[u] - cost[v]));
        }
    }
    int root = sorted.back();
    std::cout << dp[root] + cost[root] << "\n";
    return 0;
}
