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;
using ll = long long;
using pii = pair<int, int>;
using db = double;
using vi = vector<int>;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define debug(x) cout << #x <<'	' << x <<endl
const int inf = 0x3f3f3f3f;
const db eps = 1e-8;
const int mod = 1e9+7;
ll qpow(ll a, ll b){
    ll ret = 1;
    while(b){
        if(b&1)ret = ret*a%mod;
        a = a*a%mod;
        b>>=1;
    }
    return ret;
}

const int maxn = 200010;
int n, m;
ll a[maxn], b[maxn], sum[maxn], f[maxn], p[maxn], dp[maxn], vis[maxn];
int Find(int x){
	return f[x] == x?x:f[x]=Find(f[x]);
} 
vector<int>G[maxn];
bool cmp(int x, int y){
	return a[x] < a[y];
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n >> m;
	for(int i=1; i<=n; ++i){
		cin >> a[i] >> b[i];
		a[i] = max(a[i]-b[i], 0ll);
		p[i] = f[i] = i;
		sum[i] = b[i];
		dp[i] = a[i];
	}
	for(int i=0, u, v; i<m; ++i){
		cin >> u >> v;
		G[v].pb(u);
		G[u].pb(v);
	}
	sort(p+1, p+1+n, cmp);
	for(int i=1; i<=n; ++i){
		int u = p[i];
		vis[u] = 1;
		for(auto v:G[u]){
			if(vis[v]){
				int x = Find(u), y = Find(v);
				if(x!=y){
					f[y] = x;
					sum[x] += sum[y];
					dp[x] = min(dp[x], max(dp[y], a[x]-sum[y]));
				}
			}
		}
	} 
	cout << dp[Find(1)] + sum[Find(1)] <<"\n";
}
