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 <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <map>
#include <vector>
#include <set>
typedef long long ll;
using namespace std;
const int N = 1e5+5;

int n, m;
vector<int> E[N];
int A[N], B[N], pos[N], vis[N], fa[N];
long long sum[N], add[N];
inline bool cmp(int x, int y) { return A[x] < A[y]; }
inline int find(int x) { return (x == fa[x]) ? x : (fa[x] = find(fa[x])); }
int main() {
    while(~scanf("%d %d", &n, &m)) {
        for(int i = 1; i <= n; ++i) E[i].clear();
        for(int i = 1; i <= n; ++i) {
            pos[i] = i; vis[i] = 0; fa[i] = i;
            scanf("%d %d", &A[i], &B[i]);
            A[i] = max(0, A[i] - B[i]);
            add[i] = A[i];
            sum[i] = B[i];
        }

        sort(pos+1, pos+n+1, cmp);
        for(int i = 0; i < m; ++i) {
            int a, b; scanf("%d %d", &a, &b);
            E[a].push_back(b);
            E[b].push_back(a);
        }

        for(int i = 1; i <= n; ++i) {
            vis[pos[i]] = 1;
            for(int j = 0; j < E[pos[i]].size(); ++j) {
                int tt = E[pos[i]][j];
                if(vis[tt]) {
                    int f1 = find(pos[i]); int f2 = find(tt);
                    if(f1 != f2) {
                        sum[f1] += sum[f2];
                        fa[f2] = f1; 
                        add[f1] = min(add[f1], add[f2] + max(0ll, A[f1] - add[f2] - sum[f2])); 
                    }
                }
            }
        }
        printf("%lld\n", sum[pos[n]] + add[pos[n]]);
    }
    return 0;
}