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;
using LL = long long;
constexpr int MAX_N = 1E5;
constexpr LL INF = 1E18;
int n, m;
vector<int> e[MAX_N];
int a[MAX_N], b[MAX_N];
int fa[MAX_N];
int p[MAX_N];
LL f[MAX_N], s[MAX_N];
int find(int x) {
    while (fa[x] >= 0 && fa[fa[x]] >= 0)
        x = fa[x] = fa[fa[x]];
    return fa[x] >= 0 ? fa[x] : x;
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n >> m;
    for (int i = 0; i < n; ++i) {
        cin >> a[i] >> b[i];
        a[i] = max(a[i] - b[i], 0);
    }
    for (int i = 0; i < m; ++i) {
        int u, v;
        cin >> u >> v;
        --u;
        --v;
        e[u].push_back(v);
        e[v].push_back(u);
    }
    fill(fa, fa + n, -1);
    iota(p, p + n, 0);
    sort(p, p + n, [&](int i, int j) {return a[i] < a[j] || a[i] == a[j] && i < j;});
    for (int i = 0; i < n; ++i) {
        int u = p[i];
        f[u] = a[u];
        s[u] = b[u];
        for (int v : e[u]) {
            if (a[v] > a[u] || a[v] == a[u] && v > u)
                continue;
            v = find(v);
            if (u == v)
                continue;
            fa[v] = u;
            s[u] += s[v];
            f[u] = min(f[u], max((LL)a[u], f[v]) - s[v]);
        }
        f[u] += s[u];
    }
    cout << f[find(0)] << "\n";
    return 0;
}