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 cat[100005];
long long dp[2][100005];
long long d[100005];
long long sum[100005];
long long s[100005];
int main() {
  int n, m, p;
  while (scanf("%d %d %d", &n, &m, &p) != EOF) {
    d[1] = 0;
    int u;
    for (int i = 2; i <= n; i++) {
      scanf("%d", &u);
      d[i] = d[i - 1] + u;
    }
    long long tmp2;
    int tmp;
    sum[0] = 0;
    for (int i = 1; i <= m; i++) {
      scanf("%d %I64d", &tmp, &tmp2);
      cat[i] = tmp2 - d[tmp];
      sum[i] = sum[i - 1] + cat[i];
    }
    cat[0] = 0;
    sort(cat + 1, cat + m + 1);
    for (int i = 0; i <= m; i++) dp[1][i] = cat[i] * i - sum[i];
    int now, pre;
    int top;
    for (int i = 0; i < p - 1; i++) {
      top = 0;
      if ((i & 1) == 0) {
        now = 0;
        pre = 1;
      } else {
        now = 1;
        pre = 0;
      }
      s[++top] = 1;
      s[++top] = 2;
      int j = 3;
      while (j <= m) {
        while (top > 1 &&
               ((sum[j] + dp[pre][j]) - (sum[s[top]] + dp[pre][s[top]])) *
                       (s[top] - s[top - 1]) <
                   ((sum[s[top]] + dp[pre][s[top]]) -
                    (sum[s[top - 1]] + dp[pre][s[top - 1]])) *
                       (j - s[top]))
          top--;
        s[++top] = j;
        j++;
      }
      dp[now][1] = 0;
      int ptr = 1;
      for (j = 2; j <= m; j++) {
        while (ptr < top && ((sum[s[ptr + 1]] + dp[pre][s[ptr + 1]]) -
                             (sum[s[ptr]] + dp[pre][s[ptr]])) <
                                cat[j] * (s[ptr + 1] - s[ptr]))
          ptr++;
        dp[now][j] =
            dp[pre][s[ptr]] + cat[j] * (j - s[ptr]) - (sum[j] - sum[s[ptr]]);
      }
    }
    if (p % 2 == 1)
      now = 1;
    else
      now = 0;
    printf("%I64d\n", dp[now][m]);
  }
  return 0;
}
