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 <cstdio>
#include <cstdlib>
#include <cstring> 
#include <cmath> 
#include <algorithm>
#include <vector> 
using namespace std; 
inline int gi() {
    register int data = 0, w = 1;
    register char ch = 0;
    while (!isdigit(ch) && ch != '-') ch = getchar(); 
    if (ch == '-') w = -1, ch = getchar(); 
    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar(); 
    return w * data; 
} 
const int MAX_N = 1e5 + 5; 
int N, M, a[MAX_N], b[MAX_N], p[MAX_N]; 
int pa[MAX_N]; 
int getf(int x) { while (x != pa[x]) x = pa[x] = pa[pa[x]]; return x; } 
vector<int> G[MAX_N]; 
long long f[MAX_N], sum[MAX_N];
bool vis[MAX_N]; 
int main () { 
	N = gi(), M = gi();
	for (int i = 1; i <= N; i++) { 
		a[i] = gi(), b[i] = gi(); 
		a[i] = max(a[i] - b[i], 0);
		p[i] = i, pa[i] = i; 
	} 
	sort(&p[1], &p[N + 1], [](const int &l, const int &r) { return a[l] < a[r]; } ); 
	for (int i = 1; i <= M; i++) { 
		int u = gi(), v = gi(); 
		G[u].push_back(v), G[v].push_back(u); 
	} 
	for (int i = 1; i <= N; i++) { 
		vector<int> son;
		int x = p[i]; 
		vis[x] = 1, sum[x] = b[x]; 
		for (auto v : G[x]) {
			if (!vis[v] || getf(x) == getf(v)) continue; 
			son.push_back(getf(v)); 
			sum[x] += sum[getf(v)]; 
			pa[getf(v)] = getf(x); 
		} 
		f[x] = sum[x] + a[x]; 
		for (auto v : son) f[x] = min(f[x], sum[x] - sum[v] + max(1ll * a[x], f[v])); 
	} 
	printf("%lld\n", f[p[N]]); 
    return 0; 
} 