#include <bits/stdc++.h>
using namespace std;
int n, m, x, y, q, cnt, fa[15 * 100005], a[15][100005];
int Find(int x) {
  if (fa[x] == x) return x;
  fa[x] = Find(fa[x]);
  return fa[x];
}
struct segment {
  int l[15], r[15], tot;
  void _merge(segment ls, segment rs, int k) {
    for (int i = 1; i <= n; i++) {
      fa[ls.l[i]] = ls.l[i];
      fa[ls.r[i]] = ls.r[i];
      fa[rs.l[i]] = rs.l[i];
      fa[rs.r[i]] = rs.r[i];
    }
    tot = ls.tot + rs.tot;
    for (int i = 1; i <= n; i++)
      if (a[i][k] == a[i][k + 1]) {
        int al = Find(ls.r[i]), ar = Find(rs.l[i]);
        if (al != ar) {
          tot--;
          fa[al] = ar;
        }
      }
    for (int i = 1; i <= n; i++) l[i] = Find(ls.l[i]), r[i] = Find(rs.r[i]);
  }
} tr[4 * 100005];
void Build(int i, int l, int r) {
  if (l == r) {
    for (int j = 1; j <= n; j++)
      if (a[j][l] != a[j - 1][l])
        tr[i].l[j] = tr[i].r[j] = ++cnt, tr[i].tot++;
      else
        tr[i].l[j] = tr[i].r[j] = tr[i].l[j - 1];
    return;
  }
  int mid = (l + r) >> 1;
  Build(2 * i, l, mid);
  Build(2 * i + 1, mid + 1, r);
  tr[i]._merge(tr[2 * i], tr[2 * i + 1], mid);
}
segment Query(int i, int l, int r, int x, int y) {
  if (l == x && r == y) return tr[i];
  int mid = (l + r) >> 1;
  if (y <= mid)
    return Query(2 * i, l, mid, x, y);
  else if (x <= mid) {
    segment temp;
    temp._merge(Query(2 * i, l, mid, x, mid),
                Query(2 * i + 1, mid + 1, r, mid + 1, y), mid);
    return temp;
  } else
    return Query(2 * i + 1, mid + 1, r, x, y);
}
int main() {
  cin.sync_with_stdio(false);
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) cin >> a[i][j];
  Build(1, 1, m);
  for (int i = 1; i <= q; i++) {
    cin >> x >> y;
    cout << Query(1, 1, m, x, y).tot << endl;
  }
}

### Prompt

Generate a CPP solution to the following problem:

In his spare time Vladik estimates beauty of the flags.

Every flag could be represented as the matrix n × m which consists of positive integers.

Let's define the beauty of the flag as number of components in its matrix. We call component a set of cells with same numbers and between any pair of cells from that set there exists a path through adjacent cells from same component. Here is the example of the partitioning some flag matrix into components:

<image>

But this time he decided to change something in the process. Now he wants to estimate not the entire flag, but some segment. Segment of flag can be described as a submatrix of the flag matrix with opposite corners at (1, l) and (n, r), where conditions 1 ≤ l ≤ r ≤ m are satisfied.

Help Vladik to calculate the beauty for some segments of the given flag.

Input

First line contains three space-separated integers n, m, q (1 ≤ n ≤ 10, 1 ≤ m, q ≤ 105) — dimensions of flag matrix and number of segments respectively.

Each of next n lines contains m space-separated integers — description of flag matrix. All elements of flag matrix is positive integers not exceeding 106.

Each of next q lines contains two space-separated integers l, r (1 ≤ l ≤ r ≤ m) — borders of segment which beauty Vladik wants to know.

Output

For each segment print the result on the corresponding line.

Example

Input

4 5 4
1 1 1 1 1
1 2 2 3 3
1 1 1 2 5
4 4 5 5 5
1 5
2 5
1 2
4 5


Output

6
7
3
4

Note

Partitioning on components for every segment from first test case:

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, x, y, q, cnt, fa[15 * 100005], a[15][100005];
int Find(int x) {
  if (fa[x] == x) return x;
  fa[x] = Find(fa[x]);
  return fa[x];
}
struct segment {
  int l[15], r[15], tot;
  void _merge(segment ls, segment rs, int k) {
    for (int i = 1; i <= n; i++) {
      fa[ls.l[i]] = ls.l[i];
      fa[ls.r[i]] = ls.r[i];
      fa[rs.l[i]] = rs.l[i];
      fa[rs.r[i]] = rs.r[i];
    }
    tot = ls.tot + rs.tot;
    for (int i = 1; i <= n; i++)
      if (a[i][k] == a[i][k + 1]) {
        int al = Find(ls.r[i]), ar = Find(rs.l[i]);
        if (al != ar) {
          tot--;
          fa[al] = ar;
        }
      }
    for (int i = 1; i <= n; i++) l[i] = Find(ls.l[i]), r[i] = Find(rs.r[i]);
  }
} tr[4 * 100005];
void Build(int i, int l, int r) {
  if (l == r) {
    for (int j = 1; j <= n; j++)
      if (a[j][l] != a[j - 1][l])
        tr[i].l[j] = tr[i].r[j] = ++cnt, tr[i].tot++;
      else
        tr[i].l[j] = tr[i].r[j] = tr[i].l[j - 1];
    return;
  }
  int mid = (l + r) >> 1;
  Build(2 * i, l, mid);
  Build(2 * i + 1, mid + 1, r);
  tr[i]._merge(tr[2 * i], tr[2 * i + 1], mid);
}
segment Query(int i, int l, int r, int x, int y) {
  if (l == x && r == y) return tr[i];
  int mid = (l + r) >> 1;
  if (y <= mid)
    return Query(2 * i, l, mid, x, y);
  else if (x <= mid) {
    segment temp;
    temp._merge(Query(2 * i, l, mid, x, mid),
                Query(2 * i + 1, mid + 1, r, mid + 1, y), mid);
    return temp;
  } else
    return Query(2 * i + 1, mid + 1, r, x, y);
}
int main() {
  cin.sync_with_stdio(false);
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) cin >> a[i][j];
  Build(1, 1, m);
  for (int i = 1; i <= q; i++) {
    cin >> x >> y;
    cout << Query(1, 1, m, x, y).tot << endl;
  }
}

```