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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
const int N = 1e5 + 5;
typedef long long ll;
const ll INF = 0x3f3f3f3f3f3f3f3f; 
using namespace std;

int n, m, a[N], b[N], id[N], pa[N], vis[N];
ll f[N], sum[N]; 
vector<int> g[N], son; 

template < typename T >
inline T read()
{
	T 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; 
}

int getf(int x) { return pa[x] == x ? x : pa[x] = getf(pa[x]); }

bool cmp(int x, int y) { return a[x] < a[y]; }

int main()
{
	n = read <int> (), m = read <int> ();
	for(int i = 1; i <= n; i++)
	{
		a[i] = read <int> (), b[i] = read <int> ();
		a[i] = max(a[i] - b[i], 0), id[i] = pa[i] = i; 
	}
	sort(id + 1, id + n + 1, cmp);
	for(int u, v, i = 1; i <= m; i++)
		u = read <int> (), v = read <int> (), g[u].push_back(v), g[v].push_back(u);
	for(int u, i = 1; i <= n; i++)
	{
		son.clear(), u = id[i];
		vis[u] = 1, sum[u] = b[u];
		for(int v, j = 0; j < g[u].size(); j++)
		{
			v = g[u][j]; if(!vis[v] || getf(v) == getf(u)) continue;
			sum[u] += sum[getf(v)], son.push_back(getf(v)), pa[getf(v)] = getf(u); 
		}
		if(!son.size()) { f[u] = a[u] + b[u]; continue; }
		f[u] = INF; 
		for(int v, j = 0; j < son.size(); j++)
			v = son[j], f[u] = min(f[u], sum[u] - sum[v] + max(1ll * a[u], f[v])); 
	}
	printf("%lld\n", f[id[n]]); 
	return 0; 
}