Problem: The Fair Nut has two arrays a and b, consisting of n numbers. He found them so long ago that no one knows when they came to him.

The Fair Nut often changes numbers in his arrays. He also is interested in how similar a and b are after every modification.

Let's denote similarity of two arrays as the minimum number of operations to apply to make arrays equal (every operation can be applied for both arrays). If it is impossible, similarity will be equal -1.

Per one operation you can choose a subarray with length k (k is fixed), and change every element a_i, which belongs to the chosen subarray, to a_i ⊕ x (x can be chosen), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). 

Nut has already calculated the similarity of the arrays after every modification. Can you do it?

Note that you just need to calculate those values, that is you do not need to apply any operations.

Input

The first line contains three numbers n, k and q (1 ≤ k ≤ n ≤ 2 ⋅ 10^5, 0 ≤ q ≤ 2 ⋅ 10^5) — the length of the arrays, the length of the subarrays, to which the operations are applied, and the number of queries.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{14}) — elements of array a.

The third line contains n integers b_1, b_2, …, b_n (0 ≤ b_i < 2^{14}) — elements of array b.

Each of the next q lines describes query and contains string s and two integers p and v (1 ≤ p ≤ n, 0 ≤ v < 2^{14}) — array, which this query changes («a» or «b» without quotes), index of changing element and its new value.

Output

On the first line print initial similarity of arrays a and b.

On the i-th of following q lines print similarity of a and b after applying first i modifications.

Examples

Input

3 3 1
0 4 2
1 2 3
b 2 5


Output

-1
1


Input

3 2 2
1 3 2
0 0 0
a 1 0
b 1 1


Output

2
-1
2

Note

In the first sample making arrays [0, 4, 2] and [1, 2, 3] is impossible with k=3. After the modification, you can apply the operation with x=1 to the whole first array (its length is equal to k), and it will be equal to the second array.

In order to make arrays equal in the second sample before changes, you can apply operations with x=1 on subarray [1, 2] of a and with x=2 on subarray [2, 3] of b.

After all queries arrays will be equal [0, 3, 2] and [1, 0, 0]. The same operations make them equal [1, 2, 2].

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9 + 7;
const int FFTMOD = 119 << 23 | 1;
const int INF = (int)1e9 + 23111992;
const long long LINF = (long long)1e18 + 23111992;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline unsigned long long isqrt(unsigned long long k) {
  unsigned long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
mt19937 mt(chrono::high_resolution_clock::now().time_since_epoch().count());
inline int myrand() { return abs((int)mt()); }
const int maxn = 2e5 + 5;
const int magic = 500;
int n, k, q;
int a[maxn];
int b[maxn];
int c[maxn];
struct block_t {
  vector<vector<int> > blocks;
  vector<vector<int> > cnt;
  vector<int> offset;
  block_t(int n) {
    int num = (n + magic - 1) / magic;
    blocks.resize(num);
    cnt.resize(num, vector<int>(1 << 14));
    for (int i = (0); i < (num); ++i) {
      int num = min((i + 1) * magic, n) - i * magic;
      blocks[i].resize(num);
      cnt[i][0] = num;
    }
    offset.resize(num);
  }
  void upd(int d, int val) {
    for (int i = 0; (i + 1) * magic - 1 < d; i++) {
      offset[i] ^= val;
    }
    if (d % magic) {
      int i = d / magic;
      for (int j = (0); j < (int((blocks[i]).size())); ++j) {
        int& x = blocks[i][j];
        cnt[i][x]--;
        x ^= offset[i];
        cnt[i][x]++;
      }
      offset[i] = 0;
      for (int j = (0); j < (d - i * magic); ++j) {
        int& x = blocks[i][j];
        cnt[i][x]--;
        x ^= val;
        cnt[i][x]++;
      }
    }
  }
  int query(int u) {
    int i = u / magic;
    int j = u % magic;
    return blocks[i][j] ^ offset[i];
  }
  int count_zero() {
    int res = 0;
    for (int i = (0); i < (int((cnt).size())); ++i) {
      res += cnt[i][offset[i]];
    }
    return res;
  }
};
void solve_small() {
  static int g[maxn];
  vector<block_t> blocks;
  for (int i = (0); i < (k); ++i) {
    blocks.push_back(block_t((n - i) / k));
  }
  auto upd = [&](int i, int val) {
    if (!val) return;
    if (i - k + 1 >= 0) {
      int d = (i - k + 1) / k + 1;
      int r = (i + 1) % k;
      blocks[r].upd(d, val);
    }
    if (i - k >= 0) {
      int d = (i - k) / k + 1;
      int r = i % k;
      blocks[r].upd(d, val);
    }
    for (int j = i; j <= i + 1 && j < k - 1; j++) {
      g[j] ^= val;
    }
  };
  for (int i = (0); i < (n); ++i) upd(i, c[i]);
  int num = 0;
  for (int i = (0); i < (k - 1); ++i) {
    if (g[i] != blocks[i % k].query(i / k)) {
      num++;
    }
  }
  if (num) {
    cout << -1 << "\n";
  } else {
    int res = n - k + 1;
    for (int i = (0); i < (int((blocks).size())); ++i) {
      res -= blocks[i].count_zero();
    }
    cout << res << "\n";
  }
  while (q--) {
    string s;
    int u, v;
    cin >> s >> u >> v;
    u--;
    if (s == "a") {
      upd(u, a[u] ^ v);
      a[u] = v;
    } else {
      upd(u, b[u] ^ v);
      b[u] = v;
    }
    int num = 0;
    for (int i = (0); i < (k - 1); ++i) {
      if (g[i] != blocks[i % k].query(i / k)) {
        num++;
      }
    }
    if (num) {
      cout << -1 << "\n";
    } else {
      int res = n - k + 1;
      for (int i = (0); i < (int((blocks).size())); ++i) {
        res -= blocks[i].count_zero();
      }
      cout << res << "\n";
    }
  }
}
void solve_large() {
  static int f[maxn];
  static int g[maxn];
  int num1 = 0, num2 = 0;
  auto upd = [&](int i, int val) {
    if (!val) return;
    for (int j = i - k + 1; j >= 0; j -= k) {
      if (j < k - 1 && f[j] != g[j]) {
        num1--;
      }
      if (f[j]) {
        num2--;
      }
      f[j] ^= val;
      if (j < k - 1 && f[j] != g[j]) {
        num1++;
      }
      if (f[j]) {
        num2++;
      }
    }
    for (int j = i - k; j >= 0; j -= k) {
      if (j < k - 1 && f[j] != g[j]) {
        num1--;
      }
      if (f[j]) {
        num2--;
      }
      f[j] ^= val;
      if (j < k - 1 && f[j] != g[j]) {
        num1++;
      }
      if (f[j]) {
        num2++;
      }
    }
    for (int j = i; j <= i + 1 && j < k - 1; j++) {
      if (f[j] != g[j]) {
        num1--;
      }
      g[j] ^= val;
      if (f[j] != g[j]) {
        num1++;
      }
    }
  };
  for (int i = (0); i < (n); ++i) upd(i, c[i]);
  if (num1) {
    cout << -1 << "\n";
  } else {
    cout << num2 << "\n";
  }
  while (q--) {
    string s;
    int u, v;
    cin >> s >> u >> v;
    u--;
    if (s == "a") {
      upd(u, a[u] ^ v);
      a[u] = v;
    } else {
      upd(u, b[u] ^ v);
      b[u] = v;
    }
    if (num1) {
      cout << -1 << "\n";
    } else {
      cout << num2 << "\n";
    }
  }
}
void chemthan() {
  cin >> n >> k >> q;
  for (int i = (0); i < (n); ++i) cin >> a[i];
  for (int i = (0); i < (n); ++i) {
    cin >> b[i];
    c[i] = a[i] ^ b[i];
  }
  if (n < 2000 || k > magic) {
    solve_large();
  } else {
    solve_small();
  }
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  if (argc > 1) {
    assert(freopen(argv[1], "r", stdin));
  }
  if (argc > 2) {
    assert(freopen(argv[2], "wb", stdout));
  }
  chemthan();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}