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;
#define pii pair <int , int>
#define F first
#define S second
const int N = 1e5 + 5, inf = 1e9 + 1;
int A[N], B[N], par[N], C, n, m;
long long W[N], Sum;
vector <int> g[N];
priority_queue <pii, vector<pii>, greater<pii>> pq[N];

void init() {
    C = n;
    for (int v = 1; v <= n; v++) {
        par[v] = v, W[v] = B[v];
        while (!pq[v].empty()) pq[v].pop();
        for (int u : g[v]) pq[v].push({A[u] - B[u], u});
    }
}
int getpar(int v) {return par[v] = (par[v] == v ? v : getpar(par[v]));}
void Union(int u, int v) {
    u = getpar(u);
    if (u == v) return ;
    par[u] = v, W[v] += W[u], C--;
    if (pq[u].size() > pq[v].size()) swap(pq[u], pq[v]);
    while (!pq[u].empty()) pq[v].push(pq[u].top()), pq[u].pop();
}
bool check(int w) {
    init();
    for (int v = 1; v <= n; v++) if (par[v] == v && w + W[v] >= A[v]) {
        while (!pq[v].empty()) {
            int u = pq[v].top().S;
            if (A[u] - B[u] > W[v] + w) break ;
            pq[v].pop(), Union(u, v);
        }
    }
    return C == 1;
}
int32_t main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> A[i] >> B[i], Sum += B[i];
    if (n == 1) return cout << max(A[1], B[1]) << "\n", 0;
    for (int i = 0; i < m; i++) {
        int u, v; cin >> u >> v;
        g[u].push_back(v), g[v].push_back(u);
    }
    int dw = -1, up = inf;
    while (up - dw > 1) {
        int md = (up+dw)>>1;
        if (check(md)) up = md;
        else dw = md;
    }
    cout << Sum + up << "\n";


    return 0;
}
