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 LL = long long;

const int N = 100000;

struct Info
{
    LL sum, need;
};

int n, m, a[N], b[N], ord[N], par[N];
Info info[N];
std::vector<int> graph[N];

bool by_a(int i, int j)
{
    return std::max(a[i] - b[i], 0) < std::max(a[j] - b[j], 0);
}

int find(int u)
{
    if (par[u] != u) {
        par[u] = find(par[u]);
    }
    return par[u];
}

int main()
{
#ifdef LOCAL_JUDGE
    freopen("D.in", "r", stdin);
#endif
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; ++ i) {
        scanf("%d%d", a + i, b + i);
        ord[i] = i;
    }
    for (int i = 0; i < m; ++ i) {
        int u, v;
        scanf("%d%d", &u, &v);
        u --;
        v --;
        graph[u].push_back(v);
        graph[v].push_back(u);
    }
    std::sort(ord, ord + n, by_a);
    LL need = 0;
    memset(par, -1, sizeof(par));
    for (int idx = 0; idx < n; ++ idx) {
        int r = ord[idx];
        par[r] = r;
        std::vector<Info> sons;
        LL tot_sum = 0;
        for (int v : graph[r]) {
            if (~par[v] && find(r) != find(v)) {
                int u = find(v);
                sons.push_back(info[u]);
                tot_sum += info[u].sum;
                par[u] = r;
            }
        }
        need = tot_sum + std::max(a[r], b[r]);
        for (auto s : sons) {
            need = std::min(need, tot_sum - s.sum + std::max((LL)a[r], s.need + b[r]));
        }
        info[r].sum = tot_sum + b[r];
        info[r].need = need;
    }
    printf("%lld\n", need);
}