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;
int n, m, p, hd, tl, q[100010];
double d[100010], sd[100010], t[100010], st[100010], neko[100010],
    dp[100010][101];
inline double X(int i) { return i; }
inline double Y(int i, int k) { return dp[i][k] + st[i]; }
inline double slope(int i, int j, int k) {
  return (Y(i, k) - Y(j, k)) / (X(i) - X(j));
}
int main() {
  cin >> n >> m >> p;
  for (int i = 2; i <= n; i++) cin >> d[i], sd[i] = sd[i - 1] + d[i];
  for (int i = 1, h; i <= m; i++) cin >> h >> t[i], t[i] = double(t[i] - sd[h]);
  sort(t + 1, t + m + 1);
  for (int i = 1; i <= m; i++) st[i] = st[i - 1] + t[i];
  for (int i = 1; i <= m; i++) dp[i][1] = t[i] * i - st[i];
  for (int k = 2; k <= p; k++) {
    hd = tl = 1;
    for (int i = 1, j; i <= m; i++) {
      while (hd < tl && slope(q[hd], q[hd + 1], k - 1) < t[i]) hd++;
      j = q[hd], dp[i][k] = dp[j][k - 1] + t[i] * (i - j) - (st[i] - st[j]);
      while (hd < tl && slope(q[tl], i, k - 1) < slope(q[tl - 1], q[tl], k - 1))
        tl--;
      q[++tl] = i;
    }
  }
  cout << (long long)dp[m][p] << endl;
  return 0;
}
