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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long N = 1e5 + 100;
const long long K = 101;
const long long INF = 1e18;
long long d[N];
long long dp[K][N];
long long a[N], pref[N];
long long cost(long long l, long long r) {
  return a[r] * (r - l + 1) - (pref[r] - (l ? pref[l - 1] : 0));
}
void calc(long long level, long long l, long long r, long long ql,
          long long qr) {
  if (l > r) {
    return;
  }
  long long opt = -1;
  long long mid = (l + r) / 2;
  for (long long i = ql; i <= min(mid, qr); i++) {
    long long cur = cost(i, mid);
    if (i) {
      cur += dp[level - 1][i - 1];
    }
    if (dp[level][mid] > cur) {
      dp[level][mid] = cur;
      opt = i;
    }
  }
  assert(opt != -1);
  calc(level, l, mid - 1, ql, opt);
  calc(level, mid + 1, r, opt, qr);
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  for (long long i = 0; i < K; i++) {
    for (long long j = 0; j < N; j++) {
      dp[i][j] = INF;
    }
  }
  long long n, m, k;
  cin >> n >> m >> k;
  for (long long i = 1; i < n; i++) {
    cin >> d[i];
    d[i] += d[i - 1];
  }
  for (long long i = 0; i < m; i++) {
    long long pos, time;
    cin >> pos >> time;
    long long x = d[pos - 1];
    a[i] = time - x;
  }
  sort(a, a + m);
  for (long long i = 0; i < m; i++) {
    pref[i] = a[i];
    if (i) {
      pref[i] += pref[i - 1];
    }
  }
  for (long long i = 0; i < m; i++) {
    dp[0][i] = cost(0, i);
  }
  for (long long level = 1; level < k; level++) {
    calc(level, 0, m - 1, 0, m - 1);
  }
  cout << dp[k - 1][m - 1] << '\n';
}
