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;

const int N = 100010;

typedef long long LL;

int fa[N]; 

struct edge {
    int to, next;
} e[N * 2];

int head[N], ecnt;

inline void adde(int from, int to) {
    e[++ecnt] = (edge){ to, head[from] }, head[from] = ecnt;
    e[++ecnt] = (edge){ from, head[to] }, head[to] = ecnt;
}

LL dp[N], a[N], b[N], sum[N];

void dfs(int u, int f) {
    sum[u] = b[u];
    for (int i = head[u]; i; i = e[i].next)
        if (e[i].to != f) dfs(e[i].to, u), sum[u] += sum[e[i].to];
    dp[u] = sum[u] + a[u];
    for (int i = head[u]; i; i = e[i].next)
        if (e[i].to != f) dp[u] = min(dp[u], sum[u] - sum[e[i].to] + max(a[u], dp[e[i].to]));
}

int id[N], vis[N]; vector<int> G[N];

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

int main() {
    int n, m; scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%lld%lld", &a[i], &b[i]), a[i] = max(0ll, a[i] - b[i]);
    for (int i = 1; i <= n; i++) id[i] = fa[i] = i;
    for (int i = 1, a, b; i <= m; i++) scanf("%d%d", &a, &b), G[a].push_back(b), G[b].push_back(a);
    sort(id + 1, id + n + 1, [](int i, int j) { return a[i] < a[j]; }), vis[id[1]] = 1;
    for (int i = 2; i <= n; i++) {
        int u = id[i]; vis[u] = 1;
        for (auto v : G[u]) if (vis[v]) {
            if (find(u) == find(v)) continue;
            adde(find(v), u), fa[find(v)] = u;
        }
    }
    dfs(id[n], 0), printf("%lld\n", dp[id[n]]);
}