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;
long long d[100005], a[100005], s[100005], q[100005], f[100005], g[100005];
long double inter(int i, int j) {
  return ((long double)f[i] + s[i] - f[j] - s[j]) / (i - j);
}
int main() {
  cin >> n >> m >> p;
  for (int i = 2; i <= n; i++) cin >> d[i], d[i] += d[i - 1];
  for (int i = 1; i <= m; i++) {
    int h, t;
    cin >> h >> t;
    a[i] = t - d[h];
  }
  sort(a + 1, a + m + 1);
  for (int i = 1; i < m + 1; i++) s[i] = s[i - 1] + a[i];
  for (int i = 1; i < m + 1; i++) f[i] = i * a[i] - s[i];
  while (--p) {
    int head = 0, tail = 0;
    q[0] = 0;
    for (int i = 1; i < m + 1; i++) {
      while (head < tail and
             f[q[head]] + s[q[head]] - q[head] * a[i] >=
                 f[q[head + 1]] + s[q[head + 1]] - q[head + 1] * a[i])
        head++;
      g[i] = i * a[i] - s[i] + f[q[head]] + s[q[head]] - q[head] * a[i];
      while (head < tail and inter(i, q[tail]) <= inter(q[tail], q[tail - 1]))
        tail--;
      q[++tail] = i;
    }
    for (int i = 1; i < m + 1; i++) f[i] = g[i];
  }
  cout << f[m] << endl;
  return 0;
}
