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;

const int N = 1e5 + 10;
typedef pair<int, int> P;
vector<int> G[N];
P p[N];
int fa[N], a[N], b[N], vis[N], dp[N];
long long sz[N];
int F(int x) { return x == fa[x] ? x : (fa[x] = F(fa[x])); }
int main() {
    int n, m;
    cin >> n >> m;
    for(int i = 1; i <= n; i++) {
        cin >> a[i] >> b[i];
        p[i] = P(max(0, a[i] - b[i]), i);
        fa[i] = i;
        sz[i] = b[i];
    }
    for(int i = 0; i < m; i++) {
        int u, v; cin >> u >> v;
        G[u].push_back(v);
        G[v].push_back(u);
    }
    sort(p + 1, p + n + 1);
    for(int i = 1; i <= n; i++) {
        int x = p[i].second, mn = p[i].first;
        vis[x] = 1;
        for(int v : G[x]) {
            if(vis[v]) {
                int fv = F(v);
                mn = min((long long)mn, max(p[i].first - sz[fv], 1LL * dp[fv]));
            }
        }
        for(int v : G[x]) {
            if(vis[v]) {
                int fv = F(v);
                if(fv == x) continue;
                fa[fv] = x;
                sz[x] += sz[fv];
            }
        }
        dp[x] = mn;
    }
    cout << dp[F(1)] + sz[F(1)] << endl;
    return 0;
}