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;
struct line {
  long long m, y0;
  line(long long y0 = 0, long long m = 0) : y0(y0), m(m) {}
};
bool check(line a, line b, line c) {
  return (a.y0 - b.y0) * (c.m - a.m) < (a.y0 - c.y0) * (b.m - a.m);
}
int nh, pos;
line H[100001];
void update(line l) {
  while (nh >= 2 && !check(H[nh - 2], H[nh - 1], l))
    if (pos == --nh) --pos;
  H[nh++] = l;
}
inline long long eval(int id, int x) { return H[id].y0 + H[id].m * x; }
long long query(int x) {
  while (pos + 1 < nh && eval(pos, x) > eval(pos + 1, x)) ++pos;
  return eval(pos, x);
}
long long dp[101][100001], D[100001], t[100001], s[100001], a[100001],
    b[100001], c[100001], d[100001];
int main() {
  int n, m, p, i, j, x;
  cin >> n >> m >> p;
  for (i = 2; i <= n; i++) {
    cin >> x;
    D[i] = D[i - 1] + x;
  }
  for (i = 1; i <= m; i++) {
    cin >> x >> t[i];
    t[i] -= D[x];
  }
  sort(t + 1, t + m + 1);
  for (i = 1; i <= m; i++) {
    s[i] = s[i - 1] + t[i];
    a[i] = dp[1][i] = t[i] * i - s[i];
    b[i] = s[i];
    c[i] = t[i];
    d[i] = -i;
  }
  for (i = 2; i <= p; i++) {
    nh = pos = 0;
    for (j = 1; j <= m; j++) {
      update(line(dp[i - 1][j - 1] + b[j - 1], d[j - 1]));
      dp[i][j] = a[j] + query(c[j]);
    }
  }
  cout << dp[p][m] << endl;
  return 0;
}
