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
// ====================================
//   author: M_sea
//   website: https://m-sea-blog.com/
// ====================================
#include <bits/stdc++.h>
#define file(x) freopen(#x".in","r",stdin); freopen(#x".out","w",stdout)
using namespace std;
typedef long long ll;

int read() {
	int X=0,w=1; char c=getchar();
	while (c<'0'||c>'9') { if (c=='-') w=-1; c=getchar(); }
	while (c>='0'&&c<='9') X=X*10+c-'0',c=getchar();
	return X*w;
}

const int N=100000+10;

int n,m,a[N],b[N],c[N],p[N],rk[N];
vector<int> G[N],E[N];
bool cmp(int x,int y) { return c[x]<c[y]; }

int f[N];
int find(int x) { return x==f[x]?x:f[x]=find(f[x]); }

ll dp[N],sum[N];
void dfs(int u) {
	sum[u]=b[u],dp[u]=1e18;
	if (E[u].empty()) { dp[u]=max(a[u],b[u]); return; }
	for (int v:E[u]) dfs(v),sum[u]+=sum[v];
	for (int v:E[u])	
		dp[u]=min(dp[u],sum[u]-sum[v]+max(1ll*c[u],dp[v]));
}

int main() {
	n=read(),m=read();
	for (int i=1;i<=n;++i) a[i]=read(),b[i]=read();
	for (int i=1;i<=m;++i) {
		int u=read(),v=read();
		G[u].emplace_back(v),G[v].emplace_back(u);
	}
	for (int i=1;i<=n;++i) c[i]=max(0,a[i]-b[i]),p[i]=i;
	sort(p+1,p+n+1,cmp);
	for (int i=1;i<=n;++i) rk[p[i]]=f[i]=i;
	for (int i=1;i<=n;++i)
		for (int v:G[p[i]])
			if (find(v)!=p[i]&&rk[v]<i)
				E[p[i]].emplace_back(find(v)),f[find(v)]=p[i];
	dfs(p[n]); printf("%lld\n",dp[p[n]]);
	return 0;
}
// /cy/cy/cy