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 <bits/stdc++.h>

using namespace std;

int n,m;
int fa[100050];
vector<int> edge[100050],tr[100050];
int id[100050];
int a[100050],b[100050];
long long dp[100050];
long long su[100050];
bool vi[100050];

bool cmp(int A,int B)
{
	return a[A] < a[B];
}

int find_fa(int u)
{
	return fa[u] == u ? u : fa[u] = find_fa(fa[u]);
}

void dfs(int pos)
{
	for(auto v : tr[pos])
	{
		dfs(v);
		su[pos] += su[v];
	}
	su[pos] += b[pos];
	dp[pos] = su[pos] + a[pos];
	for(auto v : tr[pos])
		dp[pos] = min(dp[pos],su[pos] - su[v] + max(a[pos] * 1ll,dp[v]));
}

int main()
{
	scanf("%d%d",&n,&m);
	for(int i = 1;i <= n; ++ i)
		scanf("%d%d",&a[i],&b[i]),a[i] = max(0,a[i] - b[i]),id[i] = i;
	sort(id + 1,id + 1 + n,cmp);
	for(int i = 1;i <= m; ++ i)
	{
		int u,v; scanf("%d%d",&u,&v);
		edge[u].push_back(v);
		edge[v].push_back(u);
	}
	for(int i = 1;i <= n; ++ i) fa[i] = i;
	for(int i = 1;i <= n; ++ i)
	{
		int x = id[i];
		vi[x] = 1;
		for(auto v : edge[x])
			if(vi[v] && find_fa(v) != find_fa(x))
			{
				int f = find_fa(v);
				tr[x].push_back(f);
				fa[f] = x;
			}
	}
	dfs(id[n]);
	printf("%lld\n",dp[id[n]]);
}