Problem: Problem

There are $ N $ balls, each with its own color and value.
There are $ C $ types of ball colors from $ 1 $ to $ C $, and each color has an upper limit on the number of balls that can be selected.
Maximize the total value you get when choosing at most $ M $ balls in total.

Constraints

The input satisfies the following conditions.

* $ 1 \ leq M \ leq N \ leq 10 ^ 5 $
* $ 1 \ leq C \ leq 10 ^ 5 $
* $ 0 \ leq l_i \ leq N $
* $ 1 \ leq c_i \ leq C $
* $ 1 \ leq w_i \ leq 1000 $

Input

The input is given in the following format.


$ N $ $ M $ $ C $
$ l_1 $ $ l_2 $ ... $ l_C $
$ c_1 $ $ w_1 $
$ c_2 $ $ w_2 $
...
$ c_N $ $ w_N $


All inputs are given as integers.
$ N $, $ M $, $ C $ are given on the first line, separated by blanks.
On the second line, the upper limit of the number of balls that can be selected for the color $ i $ ($ 1 \ leq i \ leq C $) is given, separated by blanks.
The color $ c_i $ of the ball $ i $ and the value $ w_i $ ($ 1 \ leq i \ leq N $) are given in the third and subsequent lines of $ N $, separated by blanks.

Output

Output the maximum value you can get on one line.

Examples

Input

3 3 2
1 1
1 1
1 100
2 10


Output

110


Input

3 3 3
1 0 1
1 1
2 100
3 1


Output

2


Input

22 7 26
11 14 15 3 11 7 16 17 1 4 2 19 4 14 16 16 3 13 17 12 7 11 2 20 12 22
6 10
1 3
13 1
16 5
4 1
20 7
18 4
26 6
9 1
12 2
21 1
21 7
18 1
14 5
24 5
6 1
3 1
2 5
21 2
7 6
10 9
15 7


Output

52

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

//#define int long long
struct Fast {Fast(){std::cin.tie(0);ios::sync_with_stdio(false);}} fast;

/* cpp template {{{ */

/* short */
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define Fi first
#define Se second
#define ALL(v) (v).begin(), (v).end()
#define RALL(v) (v).rbegin(), (v).rend()
#define X real()
#define Y imag()

/* REPmacro */
#define REPS(i, a, n) for (ll i = (a); i < (ll)(n); ++i)
#define rep(i, n) REPS(i, 0, n)
#define REP0(i,n) for (ll i = 0; i <= (ll)(n); ++i)
#define REP1(i,n) for (ll i = 1; i <= (ll)(n); ++i)
#define RREP(i, n) REPS(i, 1, n + 1)
#define DEPS(i, a, n) for (ll i = (a); i >= (ll)(n); --i)
#define DEP(i, n) DEPS(i, n, 0)
#define EACH(i, n) for (auto&& i : n)

/* debug */
#define debug(x) cerr << x << " " << "(L:" << __LINE__ << ")" << '\n';

/* alias */
using ll = long long;
using ull = unsigned long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using pii = pair<int, int>;
using D = double;
using P = complex<D>;
using vs = vector<string>;
template <typename T> using PQ = priority_queue<T>;
template <typename T> using minPQ = priority_queue<T, vector<T>, greater<T>>;

/* const */
const int INF = 1001001001;
const ll LINF = 1001001001001001001ll;
const int MOD = 1e9 + 7;
const D EPS = 1e-9;
const int dx[] = {0, 1, 0, -1, 1, -1, 1, -1}, dy[] = {1, 0, -1, 0, 1, -1, -1, 1};

/* func */
inline bool inside(int y, int x, int H, int W) {return y >= 0 && x >= 0 && y < H && x < W;}
inline int in() {int x; cin >> x; return x;}
inline ll IN() {ll x; cin >> x; return x;}
inline vs split(const string& t, char c) {vs v; stringstream s(t); string b; while(getline(s, b, c)) v.eb(b); return v;}
template <typename T> inline bool chmin(T& a, const T& b) {if (a > b) a = b; return a > b;}
template <typename T> inline bool chmax(T& a, const T& b) {if (a < b) a = b; return a < b;}
template <typename T, typename S> inline void print(const pair<T, S>& p) {cout << p.first << " " << p.second << endl;}
template <typename T> inline void print(const T& x) {cout << x << '\n';}
template <typename T, typename S> inline void print(const vector<pair<T, S>>& v) {for (auto&& p : v) print(p);}
template <typename T> inline void print(const vector<T>& v, string s = " ") {rep(i, v.size()) cout << v[i] << (i != (ll)v.size() - 1 ? s : "\n");}
template<typename T>
istream& operator >> (istream& is, vector<T>& vec){
  for(T& x: vec) is >> x;
  //for(int i=0; i<vec.size(); i++) is >> x[i]; とかでもいいです。
  return is;
}
template<typename T>
ostream& operator << (ostream& os, vector<T>& vec){
  for(int i=0; i<vec.size(); i++){
    os << vec[i] << ( i+1 == vec.size() ? "" : " " );
  }
  return os;
}
/* }}} */

vector<long long int> dijkstra(int s,int v, vector<vector<pii>> G)
{
  //s: 始点, v: 頂点数, G: 隣接リスト
  PQ<pair<long long int,int>> wait;
  vector<long long int> result(v,INF);
  result[s]=0;
  wait.push(mp(0,s));
  while(!wait.empty())
  {
    long long int nowpoint=wait.top().Se;
    long long int nowcost=wait.top().Fi;
    wait.pop();
    if(nowcost<=result[nowpoint])
    {
      rep(i,G[nowpoint].size())
      {
        long long int nextpoint=G[nowpoint][i].Se;
        long long int nextcost= G[nowpoint][i].Fi-nowcost;
        if(nextcost<result[nextpoint])
        {
          wait.push(mp(-nextcost,nextpoint));
          result[nextpoint]=nextcost;
        }
      }
    }
  }
  return result;
}

signed main() 
{
  int n,m,c;
  cin>>n>>m>>c;
  int limit[100001];
  rep(i,c) cin>>limit[i];
  pii ball[100001];
  rep(i,n)
  {
    int cc,ww;
    cin>>cc>>ww;
    ball[i]=mp(ww,cc);
  }
  sort(ball,ball+n);
  reverse(ball,ball+n);
  ll sum=0;
  int cnt=m;
  rep(i,n)
  {
    if(cnt==0) break;
    else if(limit[ball[i].Se-1]>0)
    {
      sum+=ball[i].Fi;
      limit[ball[i].Se-1]--;
    }
    else continue;
    cnt--;
  }
  print(sum);

  return 0;
}