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>
#define rep(i, a, b) for(int i = (a); i <= (b); i++)
#define per(i, a, b) for(int i = (a); i >= (b); i--)
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
#define Size(x) ((int)x.size())
using namespace std;
typedef long long ll;
const int N = 400010;
int n, m, a[N], b[N], value[N], id[N], fa[N];
vector<int> e[N];
ll f[N], g[N]; bool vis[N];

template<typename T> void gi(T &x) {
	x = 0; register char c = getchar();
	for(; c < '0' || c > '9'; c = getchar());
	for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + (c ^ 48);
}
inline bool cmp(const int &x, const int &y) { return value[x] < value[y]; }
int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
int main() {
	gi(n), gi(m);
	rep(i, 1, n) gi(a[i]), gi(b[i]), value[i] = max(0, a[i] - b[i]), id[i] = fa[i] = i;
	sort(id + 1, id + n + 1, cmp);
	rep(i, 1, m) { int x, y; gi(x), gi(y), e[x].pb(y), e[y].pb(x); }
	rep(i, 1, n) {
		vector<int> edge; int u = id[i]; vis[u] = 1, g[u] = b[u];
		for(auto v : e[u]) if(vis[v] && find(u) != find(v))
			edge.pb(v), edge.pb(find(v)), g[u] += g[find(v)], fa[find(v)] = u;
		f[u] = g[u] + value[u];
		for(auto v : edge) f[u] = min(f[u], g[u] - g[v] + max(1ll * value[u], f[v]));
	}
	printf("%lld\n", f[id[n]]);
	return 0;
}