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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <utility>
#include <cstdio>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <cmath>
#include <set>
#include <map>
#define my_abs(x) ((x) < 0 ? -(x) : (x))
#define mp std::make_pair
#define pb push_back
typedef long long ll; 
int a[100005], b[100005], fa[100005], idx[100005]; 
ll dp[100005], sum[100005]; 
std::vector<int> adj[100005], son; 
bool vis[100005]; 
int getfa(int u)
{
	if (u == fa[u])
		return u; 
	return fa[u] = getfa(fa[u]); 
}
int main()
{
	// freopen("ARC098-F.in", "r", stdin); 
	int n, m; 
	scanf("%d%d", &n, &m); 
	for (int i = 0; i < n; i++)
	{
		scanf("%d%d", a + i, b + i); 
		a[i] = std::max(a[i] - b[i], 0); 
		fa[i] = idx[i] = i; 
	}
	std::sort(idx, idx + n, [&] (int x, int y) { return a[x] < a[y]; }); 
	for (int i = 0; i < m; i++)
	{
		int u, v; 
		scanf("%d%d", &u, &v); 
		adj[--u].pb(--v); 
		adj[v].pb(u); 
	}
	for (int i = 0; i < n; i++)
	{
		int u = idx[i]; 
		vis[u] = true; 
		son.clear(); 
		sum[u] = b[u]; 
		for (int v : adj[u])
		{
			if (!vis[v] || getfa(v) == getfa(u))
				continue; 
			sum[u] += sum[getfa(v)]; 
			son.pb(getfa(v)); 
			fa[getfa(v)] = getfa(u); 
		}
		dp[u] = sum[u] + a[u]; 
		for (int v : son)
			dp[u] = std::min(dp[u], sum[u] - sum[v] + std::max((ll)a[u], dp[v])); 
	}
	printf("%lld\n", dp[idx[n - 1]]);
	return 0; 
}
