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 <cstdio>
#include <algorithm>
#include <vector>
#include <iostream>
using namespace std;

const int N = 100010;
typedef long long ll;
int A[N] , B[N];

int n , m;
vector<int> adj[N] , son[N];

struct Cmp {  bool operator() (int x , int y) const {return A[x] < A[y];} };

int par[N] , id[N] , vis[N];
int find(int x) {return x == par[x] ? x : par[x] = find(par[x]);}

ll dp[N] , s[N];
void dfs(int x) {
    s[x] = B[x];
    for(auto y : son[x]) dfs(y) , s[x] += s[y];
    dp[x] = A[x] + s[x];
    for(auto y : son[x]) dp[x] = min(dp[x] , max(dp[y] , (ll)A[x]) + s[x] - s[y]);
}
int main() {
    scanf("%d %d" , &n , &m);
    for(int i = 1 ; i <= n ; ++i) {
        scanf("%d %d" , A + i , B + i);
        A[i] = max(A[i] - B[i] , 0);
        par[i] = i , id[i] = i;
    }
    sort(id + 1 , id + 1 + n , Cmp());
    for(int i = 1 , x , y; i <= m ; ++i) {
        scanf("%d %d" , &x , &y);
        adj[x].push_back(y) , adj[y].push_back(x);
    }
    for(int i = 1 ; i <= n ; ++i) {
        int x = id[i];
        vis[x] = 1;
        for(auto y : adj[x])
            if(vis[y]) {
                int z = find(y);
                if(z != x) son[x].push_back(z) , par[z] = x;
            }
    }
    dfs(id[n]) , cout << dp[id[n]] << endl;
}