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;
long long d[111111];
long long sum[111111];
int q[111111];
long long cat[111111];
long long dp[101][100001];
long long getDP(int i, int j, int k) {
  return (long long)j * cat[j] + dp[i - 1][k] - j * cat[k];
}
long long getUP(int i, int j, int k) {
  return (long long)dp[i - 1][j] - dp[i - 1][k];
}
long long getDOWN(int j, int k) { return (long long)cat[j] - cat[k]; }
int main() {
  int i, j, k, n, m, p, h;
  cin >> n >> m >> p;
  for (i = 2; i <= n; i++) {
    cin >> d[i];
    d[i] += d[i - 1];
  }
  for (i = 1; i <= m; i++) {
    cin >> h >> cat[i];
    cat[i] -= d[h];
  }
  sort(cat + 1, cat + 1 + m);
  for (i = 1; i <= m; i++) sum[i] = sum[i - 1] + cat[i];
  for (i = 1; i <= p; i++)
    for (j = 1; j <= m; j++) dp[i][j] = 999999999999999;
  dp[1][m] = (long long)m * cat[m] - sum[m];
  for (i = 2; i <= p; i++) {
    int head, tail;
    head = tail = 0;
    q[tail++] = m;
    for (j = m; j >= 1; j--) {
      while (head + 1 < tail && (getUP(i, q[head + 1], q[head]) <=
                                 j * getDOWN(q[head + 1], q[head])))
        head++;
      dp[i][j] = min(dp[i][j], getDP(i, j, q[head]));
      while (head + 1 < tail &&
             getUP(i, j, q[tail - 1]) * getDOWN(q[tail - 1], q[tail - 2]) >
                 getUP(i, q[tail - 1], q[tail - 2]) * getDOWN(j, q[tail - 1]))
        tail--;
      q[tail++] = j;
    }
  }
  long long ans = dp[p][1];
  for (i = 1; i <= m; i++) ans = min(ans, dp[p][i]);
  cout << ans << endl;
}
