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 N = 1e5 + 200;
const int M = 1e5 + 100;
template <class Ruby_Rose>
inline void read(Ruby_Rose &x) {
  x = 0;
  char ch = getchar(), w = 0;
  while (!isdigit(ch)) w = (ch == '-'), ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  x = w ? -x : x;
}
int n, m, p, q[M];
long long dis[N], val[M], f[M][2], sum[N];
inline int check(int a, int b, int c, int d) {
  return 1.0 * (f[c][d] + sum[c] - f[b][d] - sum[b]) / (c - b) <=
         1.0 * (f[b][d] + sum[b] - f[a][d] - sum[a]) / (b - a);
}
int main() {
  read(n), read(m), read(p);
  for (register int i = 2; i <= n; i++) read(dis[i]), dis[i] += dis[i - 1];
  for (register int i = 1, h; i <= m; i++)
    read(h), read(val[i]), val[i] -= dis[h];
  sort(val + 1, val + 1 + m);
  for (register int i = 1; i <= m; i++) sum[i] = sum[i - 1] + val[i];
  f[0][0] = 0;
  for (register int i = 1; i <= m; i++) f[i][0] = 1e15;
  for (register int j = 1; j <= p; j++) {
    int st = 1, ed = 0;
    for (register int i = 1; i <= m; i++) {
      while (st < ed && check(q[ed - 1], q[ed], i - 1, j & 1 ^ 1)) ed--;
      q[++ed] = i - 1;
      while (st < ed && 1.0 *
                                (f[q[st + 1]][j & 1 ^ 1] + sum[q[st + 1]] -
                                 f[q[st]][j & 1 ^ 1] - sum[q[st]]) /
                                (q[st + 1] - q[st]) <=
                            1.0 * val[i])
        st++;
      f[i][j & 1] =
          f[q[st]][j & 1 ^ 1] + (i - q[st]) * val[i] - (sum[i] - sum[q[st]]);
    }
  }
  cout << f[m][p & 1] << endl;
  return 0;
}
