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>
using namespace std;
const int INF = 1e9 + 1;
const long long INFLL = 2e18 + 1;
vector<long long> pr;
long long get(int l, int r) { return pr[r + 1] - pr[l]; }
vector<vector<long long>> dp;
vector<long long> a;
void solve(int l, int r, int optl, int optr, int k) {
  if (l > r) return;
  int m = (l + r) / 2;
  long long ans = INFLL;
  int optm = -1;
  for (int j = max(0, optl); j <= optr && j <= m; j++) {
    if (ans > dp[j][k - 1] + a[m] * (m - j + 1) - get(j, m)) {
      ans = dp[j][k - 1] + a[m] * (m - j + 1) - get(j, m);
      optm = j;
    }
  }
  dp[m + 1][k] = ans;
  solve(l, m - 1, optl, optm, k);
  solve(m + 1, r, optm, optr, k);
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  int n, m, p;
  cin >> n >> m >> p;
  p = min(p, m);
  vector<long long> d(n);
  for (int i = 0; i < n - 1; i++) {
    int x;
    cin >> x;
    d[i + 1] = d[i] + x;
  }
  a.resize(m);
  for (int i = 0; i < m; i++) {
    int h, t;
    cin >> h >> t;
    a[i] = t - d[h - 1];
  }
  sort((a).begin(), (a).end());
  pr.resize(m + 1);
  for (int i = 0; i < m; i++) pr[i + 1] = pr[i] + a[i];
  dp.assign(m + 1, vector<long long>(p + 1, INFLL));
  dp[0][0] = 0;
  for (int k = 1; k <= p; k++) {
    solve(0, m - 1, 0, m - 1, k);
  }
  cout << dp[m][p];
  return 0;
}
