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 <cstring>
#include <algorithm>
#include <vector>
#define MN 101000

typedef long long ll;

int h[MN], nxt[2 * MN], to[2 * MN], K = 0;
int n, m, fa[MN]; 
int a[MN], c[MN], p[MN];
std::vector<int> e[MN];
ll sum[MN];
bool vis[MN];
ll Ans = 1e18;

void ins(int u, int v) {nxt[++K] = h[u]; h[u] = K; to[K] = v;} 

int Find(int x) {return fa[x] == x ? x : fa[x] = Find(fa[x]);}

bool cmp(int a, int b) {return c[a] < c[b];}

void dfs(int x)
{
	sum[x] = a[x] - c[x];
	for(int i = h[x]; i; i = nxt[i])
		dfs(to[i]), sum[x] += sum[to[i]];
}

void solve(int x, ll w)
{
	Ans = std::min(Ans, std::max((ll)c[x], w - sum[x]));
	for(int i = h[x]; i; i = nxt[i])
	{
		ll d = std::max(w - (sum[x] - sum[to[i]]), (ll)c[x]);
		solve(to[i], d);
	}
}

int main()
{
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i++)
	{
		int a, b; scanf("%d%d", &a, &b);
		a = std::max(a, b);
		::a[i] = a; c[i] = a - b;
		fa[i] = i; p[i] = i;
	}
	for(int i = 1; i <= m; i++)
	{
		int u, v; scanf("%d%d", &u, &v);
		e[u].push_back(v);
		e[v].push_back(u);
	}

	std::sort(p + 1, p + n + 1, cmp);
	
	for(int i = 1; i <= n; i++)
	{
		int x = p[i]; vis[x] = 1;
		for(int j = 0; j < e[x].size(); j++)
		{
			if(vis[e[x][j]])
			{
				int y = Find(e[x][j]);
				if(x != y)
				{
					ins(x, y);
					fa[y] = x;
				}
			}
		}
	}
	
	dfs(p[n]);
	solve(p[n], 0);
	
	printf("%lld\n", Ans + sum[p[n]]); 
}