Zxr960115 is owner of a large farm. He feeds m cute cats and employs p feeders. There's a straight road across the farm and n hills along the road, numbered from 1 to n from left to right. The distance between hill i and (i - 1) is di meters. The feeders live in hill 1.

One day, the cats went out to play. Cat i went on a trip to hill hi, finished its trip at time ti, and then waited at hill hi for a feeder. The feeders must take all the cats. Each feeder goes straightly from hill 1 to n without waiting at a hill and takes all the waiting cats at each hill away. Feeders walk at a speed of 1 meter per unit time and are strong enough to take as many cats as they want.

For example, suppose we have two hills (d2 = 1) and one cat that finished its trip at time 3 at hill 2 (h1 = 2). Then if the feeder leaves hill 1 at time 2 or at time 3, he can take this cat, but if he leaves hill 1 at time 1 he can't take it. If the feeder leaves hill 1 at time 2, the cat waits him for 0 time units, if the feeder leaves hill 1 at time 3, the cat waits him for 1 time units.

Your task is to schedule the time leaving from hill 1 for each feeder so that the sum of the waiting time of all cats is minimized.

Input

The first line of the input contains three integers n, m, p (2 ≤ n ≤ 105, 1 ≤ m ≤ 105, 1 ≤ p ≤ 100).

The second line contains n - 1 positive integers d2, d3, ..., dn (1 ≤ di < 104).

Each of the next m lines contains two integers hi and ti (1 ≤ hi ≤ n, 0 ≤ ti ≤ 109).

Output

Output an integer, the minimum sum of waiting time of all cats.

Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

4 6 2
1 3 5
1 0
2 1
4 9
1 10
2 10
3 12


Output

3
#include <bits/stdc++.h>
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const double EPS = 0.000000000000001;
const double PI = acos(-1.0);
const int MOD = 1e9 + 7;
using namespace std;
void fast_io() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
void file_in(const string& filename) { freopen(filename.c_str(), "r", stdin); }
void file_out(const string& filename) {
  freopen(filename.c_str(), "w", stdout);
}
double diffclock(clock_t clock1, clock_t clock2) {
  double diffticks = clock2 - clock1;
  return diffticks / ((double)CLOCKS_PER_SEC / 1000);
}
void sleep(int milliseconds) {
  clock_t start = clock();
  for (clock_t end = clock(); diffclock(start, end) < milliseconds;
       end = clock()) {
  }
}
struct Line {
  long long k;
  long long b;
  Line() : Line(0, 0) {}
  Line(long long k, long long b) : k(k), b(b) {}
  long long get(int x) const { return k * x + b; }
};
long long cross(Line first, Line second) {
  return (second.b - first.b - 1) / (first.k - second.k) + 1;
}
void add_line(Line new_line, vector<Line>& lines, vector<long long>& points) {
  while (!points.empty() && (cross(lines.back(), new_line) < points.back())) {
    points.pop_back();
    lines.pop_back();
  }
  if (points.empty()) {
    points.push_back(-LINF);
  } else {
    points.push_back(cross(lines.back(), new_line));
  }
  lines.push_back(new_line);
}
long long get_value(int x, vector<Line>& lines, vector<long long>& points,
                    int& last_index) {
  while ((last_index + 1 < points.size()) && (points[last_index + 1] <= x)) {
    ++last_index;
  }
  return lines[last_index].get(x);
}
void solve() {
  int n, m, p;
  cin >> n >> m >> p;
  vector<int> dist(n);
  for (int i = 1; i < n; ++i) {
    int cur_dist;
    cin >> cur_dist;
    dist[i] = dist[i - 1] + cur_dist;
  }
  vector<int> start_time(m + 1, -2 * INF);
  int min_value = 2 * INF;
  int max_value = -2 * INF;
  for (int i = 0; i < m; ++i) {
    int h, t;
    cin >> h >> t;
    start_time[i] = t - dist[h - 1];
  }
  sort(start_time.begin(), start_time.end());
  vector<long long> prefix(m + 1);
  for (int i = 1; i <= m; ++i) {
    prefix[i] = prefix[i - 1] - start_time[i];
  }
  vector<vector<long long> > dp(m + 1, vector<long long>(p + 1, LINF));
  for (int i = 0; i <= p; ++i) {
    dp[0][i] = 0;
  }
  vector<vector<Line> > lines(p + 1);
  vector<vector<long long> > points(p + 1);
  vector<int> last_index(p + 1);
  add_line(Line(-1, 0), lines[0], points[0]);
  for (int i = 1; i <= m; ++i) {
    for (int j = 1; j <= p; ++j) {
      dp[i][j] = get_value(start_time[i], lines[j - 1], points[j - 1],
                           last_index[j - 1]) +
                 (long long)start_time[i] * (i + 1) + prefix[i];
      add_line(Line(-(i + 1), dp[i][j] - prefix[i]), lines[j], points[j]);
    }
  }
  cout << dp[m][p] << '\n';
}
signed main() {
  fast_io();
  solve();
  return 0;
}
