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 MAXN = 100010;
const int MAXM = 110;
long long n, m, p;
long long d[MAXN];
long long h[MAXN], t[MAXN], s[MAXN], a[MAXN];
long long dp[MAXM][MAXN], g[MAXN];
int q[MAXN];
long long calc(int j, int k) { return g[j] + s[j] - g[k] - s[k]; }
int main() {
  scanf("%I64d%I64d%I64d", &n, &m, &p);
  if (p >= m) {
    cout << 0 << endl;
    return 0;
  }
  d[1] = 0LL;
  for (int i = 2; i <= n; i++) {
    scanf("%I64d", &d[i]);
    d[i] += d[i - 1];
  }
  for (int i = 1; i <= m; i++) {
    scanf("%I64d%I64d", &h[i], &t[i]);
    a[i] = t[i] - d[h[i]];
  }
  sort(a + 1, a + m + 1);
  s[0] = 0;
  for (int i = 1; i <= m; i++) s[i] = s[i - 1] + a[i];
  for (int i = 1; i <= m; i++) dp[1][i] = i * a[i] - s[i];
  for (int t = 2; t <= p; t++) {
    int l = 1, r = 0;
    for (int i = 1; i <= m; i++) g[i] = dp[t - 1][i];
    g[0] = 0;
    q[++r] = 0;
    for (int i = 1; i <= m; i++) {
      while (l < r && calc(q[l + 1], q[l]) <= a[i] * (q[l + 1] - q[l])) l++;
      int j = q[l];
      dp[t][i] = dp[t - 1][j] + (i - j) * a[i] - (s[i] - s[j]);
      while (l < r && calc(q[r], q[r - 1]) * (i - q[r]) >=
                          calc(i, q[r]) * (q[r] - q[r - 1]))
        r--;
      q[++r] = i;
    }
  }
  cout << dp[p][m] << endl;
}
