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;
#define pb push_back
typedef long long ll;
int n, m;
const int maxn = 1e5 + 10;
int a[maxn], b[maxn], c[maxn], fa[maxn];
ll sumb[maxn], dp[maxn];
vector<int> ver, G[maxn], E[maxn];
bool cmp(const int & i, const int & j) { return c[i] < c[j] || (c[i] == c[j] && i < j);}
int getfa(int x) { return fa[x] == x ? x : fa[x] = getfa(fa[x]);}
void dfs(int x)
{
	sumb[x] = b[x];
	for(auto v : E[x])
	{
		dfs(v);
		sumb[x] += sumb[v];
	}
	dp[x] = sumb[x] + c[x];
	for(auto v : E[x])
		dp[x] = min(dp[x], sumb[x] - sumb[v] + max((ll)c[x], dp[v]));
}
int main()
{
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; ++i)
	{
		scanf("%d%d", &a[i], &b[i]);
		c[i] = max(a[i] - b[i], 0);
		ver.push_back(i); 
		fa[i] = i;
	}
	for(int i = 1; i <= m; ++i)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		G[u].pb(v);
		G[v].pb(u);
	}
	sort(ver.begin(), ver.end(), cmp);
	for(auto u : ver)
		for(auto v : G[u])
		{
			int fv = getfa(v);
			if(c[fv] > c[u] || (c[fv] == c[u] && fv >= u))
				continue;
			fa[fv] = u;
			E[u].push_back(fv);
		}
	int rt = ver.back();
	dfs(rt);
	printf("%lld\n", dp[rt]);
	return 0;
}
	
	 